142083
The surface of zone material is radiated in turn by waves of $\lambda=350 \mathrm{~nm}$ and $540 \mathrm{~nm}$ respectively. The ratio of the stopping potential in the two cases is $2: 1$. The work function of the material is
1 $4.20 \mathrm{eV}$
2 $0.15 \mathrm{eV}$
3 $2.10 \mathrm{eV}$
4 $1.05 \mathrm{eV}$
Explanation:
D Given that, $\lambda_{1}=350 \mathrm{~nm}, \mathrm{~V}_{\mathrm{s}_{1}}=2 \mathrm{~V}_{\mathrm{s}_{2}}, \lambda_{2}=540 \mathrm{~nm}$ According to Einstein photoelectric equation - $\frac{\mathrm{hc}}{\lambda}=\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}$ $\lambda \propto \frac{1}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{1}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $\frac{540}{350}=\frac{\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $54\left(\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}\right)=35\left(\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}\right)$ $\mathrm{V}_{\mathrm{s}_{2}}=\frac{19}{16} \phi_{\mathrm{o}}$ From equation (i), we get - $\phi_{\mathrm{o}} =\frac{\mathrm{hc}}{\lambda_{1}}-\mathrm{V}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-2 \mathrm{~V}_{\mathrm{s}_{2}}$ $\phi_{\mathrm{o}} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{-9} \times 1.6 \times 10^{-19}}-2 \times \frac{19}{16} \phi_{\mathrm{o}}$ $\phi_{\mathrm{o}} =0.0105 \times 10^{2}$ $\phi_{\mathrm{o}} =1.05 \mathrm{eV}$
AIIMS-2000
Dual nature of radiation and Matter
142084
In a photo emissive cell with executing wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of fastest emitted electron will be-
1 greater than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
2 $\mathrm{v}\left(\frac{3}{4}\right)^{1 / 2}$
3 $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
4 less than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
Explanation:
A From Einstein photoelectric equation $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-\lambda\right)}{\lambda \lambda_{0}}}$ If wavelength is changed to $\left(\frac{3 \lambda}{4}\right)$ $\frac{1}{2} \mathrm{mv}^{\prime 2}=\frac{4 h \mathrm{c}}{3 \lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\lambda_{0}(3 \lambda / 4)}}$ Form equation (i) and (ii), we get- $\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\sqrt{\frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\frac{3}{4} \lambda \lambda_{0}} \times \frac{\lambda \lambda_{0}}{\lambda_{0}-\lambda}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{4}{3} \times \mathrm{v} \times \sqrt{\frac{\lambda_{0}-3 \lambda / 4}{\lambda_{0}-\lambda}}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3} \Rightarrow \mathrm{v}^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \mathrm{v}$
BCECE-2015
Dual nature of radiation and Matter
142086
A radio transmitter operates at a frequency of $880 \mathrm{kHz}$ and power of $10 \mathrm{~kW}$. The number of photons emitted per second is-
1 $13.27 \times 10^{4}$
2 $13.27 \times 10^{34}$
3 $1327 \times 10^{34}$
4 $1.71 \times 10^{31}$
Explanation:
D Given that, Frequency $(v)=880 \mathrm{kHz}$ Power $(\mathrm{P})=10 \mathrm{~kW}$ Number of photons emitted per second, $\mathrm{h}=\frac{\text { Power }}{\text { Energy of photon }}$ $\mathrm{h}=\frac{\text { Power }}{\mathrm{h} v}$ $\mathrm{~h}=\frac{10}{6.6 \times 10^{-34} \times 880}$ $\mathrm{~h}=1.72 \times 10^{31}$
BCECE-2011
Dual nature of radiation and Matter
142080
Assertion: Photoelectric saturation current increases with the increase in frequency of incident light. Reason: Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
D Photoelectric saturation current is independent of frequency. It only depends on the intensity of light.
AIIMS-2015
Dual nature of radiation and Matter
142087
Light of wavelength $5000 \AA$ falling on a sensitive surface. If the surface has received $10^{-}$ ${ }^{7} \mathbf{J}$ of energy, then the number of photons falling on the surface will be-
1 $5 \times 10^{11}$
2 $2.5 \times 10^{11}$
3 $3 \times 10^{11}$
4 none of these
Explanation:
B Given that, $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$ $\mathrm{E}=10^{-7} \mathrm{~J}$ $\therefore$ Number of photons $(n)=\frac{E}{h c / \lambda}=\frac{E \lambda}{h c}$ $\mathrm{n}=\frac{10^{-7} \times 5 \times 10^{7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=252 \times 10^{9}$ $\mathrm{n} \approx 2.5 \times 10^{11}$
142083
The surface of zone material is radiated in turn by waves of $\lambda=350 \mathrm{~nm}$ and $540 \mathrm{~nm}$ respectively. The ratio of the stopping potential in the two cases is $2: 1$. The work function of the material is
1 $4.20 \mathrm{eV}$
2 $0.15 \mathrm{eV}$
3 $2.10 \mathrm{eV}$
4 $1.05 \mathrm{eV}$
Explanation:
D Given that, $\lambda_{1}=350 \mathrm{~nm}, \mathrm{~V}_{\mathrm{s}_{1}}=2 \mathrm{~V}_{\mathrm{s}_{2}}, \lambda_{2}=540 \mathrm{~nm}$ According to Einstein photoelectric equation - $\frac{\mathrm{hc}}{\lambda}=\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}$ $\lambda \propto \frac{1}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{1}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $\frac{540}{350}=\frac{\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $54\left(\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}\right)=35\left(\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}\right)$ $\mathrm{V}_{\mathrm{s}_{2}}=\frac{19}{16} \phi_{\mathrm{o}}$ From equation (i), we get - $\phi_{\mathrm{o}} =\frac{\mathrm{hc}}{\lambda_{1}}-\mathrm{V}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-2 \mathrm{~V}_{\mathrm{s}_{2}}$ $\phi_{\mathrm{o}} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{-9} \times 1.6 \times 10^{-19}}-2 \times \frac{19}{16} \phi_{\mathrm{o}}$ $\phi_{\mathrm{o}} =0.0105 \times 10^{2}$ $\phi_{\mathrm{o}} =1.05 \mathrm{eV}$
AIIMS-2000
Dual nature of radiation and Matter
142084
In a photo emissive cell with executing wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of fastest emitted electron will be-
1 greater than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
2 $\mathrm{v}\left(\frac{3}{4}\right)^{1 / 2}$
3 $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
4 less than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
Explanation:
A From Einstein photoelectric equation $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-\lambda\right)}{\lambda \lambda_{0}}}$ If wavelength is changed to $\left(\frac{3 \lambda}{4}\right)$ $\frac{1}{2} \mathrm{mv}^{\prime 2}=\frac{4 h \mathrm{c}}{3 \lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\lambda_{0}(3 \lambda / 4)}}$ Form equation (i) and (ii), we get- $\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\sqrt{\frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\frac{3}{4} \lambda \lambda_{0}} \times \frac{\lambda \lambda_{0}}{\lambda_{0}-\lambda}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{4}{3} \times \mathrm{v} \times \sqrt{\frac{\lambda_{0}-3 \lambda / 4}{\lambda_{0}-\lambda}}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3} \Rightarrow \mathrm{v}^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \mathrm{v}$
BCECE-2015
Dual nature of radiation and Matter
142086
A radio transmitter operates at a frequency of $880 \mathrm{kHz}$ and power of $10 \mathrm{~kW}$. The number of photons emitted per second is-
1 $13.27 \times 10^{4}$
2 $13.27 \times 10^{34}$
3 $1327 \times 10^{34}$
4 $1.71 \times 10^{31}$
Explanation:
D Given that, Frequency $(v)=880 \mathrm{kHz}$ Power $(\mathrm{P})=10 \mathrm{~kW}$ Number of photons emitted per second, $\mathrm{h}=\frac{\text { Power }}{\text { Energy of photon }}$ $\mathrm{h}=\frac{\text { Power }}{\mathrm{h} v}$ $\mathrm{~h}=\frac{10}{6.6 \times 10^{-34} \times 880}$ $\mathrm{~h}=1.72 \times 10^{31}$
BCECE-2011
Dual nature of radiation and Matter
142080
Assertion: Photoelectric saturation current increases with the increase in frequency of incident light. Reason: Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
D Photoelectric saturation current is independent of frequency. It only depends on the intensity of light.
AIIMS-2015
Dual nature of radiation and Matter
142087
Light of wavelength $5000 \AA$ falling on a sensitive surface. If the surface has received $10^{-}$ ${ }^{7} \mathbf{J}$ of energy, then the number of photons falling on the surface will be-
1 $5 \times 10^{11}$
2 $2.5 \times 10^{11}$
3 $3 \times 10^{11}$
4 none of these
Explanation:
B Given that, $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$ $\mathrm{E}=10^{-7} \mathrm{~J}$ $\therefore$ Number of photons $(n)=\frac{E}{h c / \lambda}=\frac{E \lambda}{h c}$ $\mathrm{n}=\frac{10^{-7} \times 5 \times 10^{7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=252 \times 10^{9}$ $\mathrm{n} \approx 2.5 \times 10^{11}$
142083
The surface of zone material is radiated in turn by waves of $\lambda=350 \mathrm{~nm}$ and $540 \mathrm{~nm}$ respectively. The ratio of the stopping potential in the two cases is $2: 1$. The work function of the material is
1 $4.20 \mathrm{eV}$
2 $0.15 \mathrm{eV}$
3 $2.10 \mathrm{eV}$
4 $1.05 \mathrm{eV}$
Explanation:
D Given that, $\lambda_{1}=350 \mathrm{~nm}, \mathrm{~V}_{\mathrm{s}_{1}}=2 \mathrm{~V}_{\mathrm{s}_{2}}, \lambda_{2}=540 \mathrm{~nm}$ According to Einstein photoelectric equation - $\frac{\mathrm{hc}}{\lambda}=\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}$ $\lambda \propto \frac{1}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{1}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $\frac{540}{350}=\frac{\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $54\left(\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}\right)=35\left(\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}\right)$ $\mathrm{V}_{\mathrm{s}_{2}}=\frac{19}{16} \phi_{\mathrm{o}}$ From equation (i), we get - $\phi_{\mathrm{o}} =\frac{\mathrm{hc}}{\lambda_{1}}-\mathrm{V}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-2 \mathrm{~V}_{\mathrm{s}_{2}}$ $\phi_{\mathrm{o}} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{-9} \times 1.6 \times 10^{-19}}-2 \times \frac{19}{16} \phi_{\mathrm{o}}$ $\phi_{\mathrm{o}} =0.0105 \times 10^{2}$ $\phi_{\mathrm{o}} =1.05 \mathrm{eV}$
AIIMS-2000
Dual nature of radiation and Matter
142084
In a photo emissive cell with executing wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of fastest emitted electron will be-
1 greater than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
2 $\mathrm{v}\left(\frac{3}{4}\right)^{1 / 2}$
3 $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
4 less than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
Explanation:
A From Einstein photoelectric equation $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-\lambda\right)}{\lambda \lambda_{0}}}$ If wavelength is changed to $\left(\frac{3 \lambda}{4}\right)$ $\frac{1}{2} \mathrm{mv}^{\prime 2}=\frac{4 h \mathrm{c}}{3 \lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\lambda_{0}(3 \lambda / 4)}}$ Form equation (i) and (ii), we get- $\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\sqrt{\frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\frac{3}{4} \lambda \lambda_{0}} \times \frac{\lambda \lambda_{0}}{\lambda_{0}-\lambda}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{4}{3} \times \mathrm{v} \times \sqrt{\frac{\lambda_{0}-3 \lambda / 4}{\lambda_{0}-\lambda}}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3} \Rightarrow \mathrm{v}^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \mathrm{v}$
BCECE-2015
Dual nature of radiation and Matter
142086
A radio transmitter operates at a frequency of $880 \mathrm{kHz}$ and power of $10 \mathrm{~kW}$. The number of photons emitted per second is-
1 $13.27 \times 10^{4}$
2 $13.27 \times 10^{34}$
3 $1327 \times 10^{34}$
4 $1.71 \times 10^{31}$
Explanation:
D Given that, Frequency $(v)=880 \mathrm{kHz}$ Power $(\mathrm{P})=10 \mathrm{~kW}$ Number of photons emitted per second, $\mathrm{h}=\frac{\text { Power }}{\text { Energy of photon }}$ $\mathrm{h}=\frac{\text { Power }}{\mathrm{h} v}$ $\mathrm{~h}=\frac{10}{6.6 \times 10^{-34} \times 880}$ $\mathrm{~h}=1.72 \times 10^{31}$
BCECE-2011
Dual nature of radiation and Matter
142080
Assertion: Photoelectric saturation current increases with the increase in frequency of incident light. Reason: Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
D Photoelectric saturation current is independent of frequency. It only depends on the intensity of light.
AIIMS-2015
Dual nature of radiation and Matter
142087
Light of wavelength $5000 \AA$ falling on a sensitive surface. If the surface has received $10^{-}$ ${ }^{7} \mathbf{J}$ of energy, then the number of photons falling on the surface will be-
1 $5 \times 10^{11}$
2 $2.5 \times 10^{11}$
3 $3 \times 10^{11}$
4 none of these
Explanation:
B Given that, $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$ $\mathrm{E}=10^{-7} \mathrm{~J}$ $\therefore$ Number of photons $(n)=\frac{E}{h c / \lambda}=\frac{E \lambda}{h c}$ $\mathrm{n}=\frac{10^{-7} \times 5 \times 10^{7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=252 \times 10^{9}$ $\mathrm{n} \approx 2.5 \times 10^{11}$
142083
The surface of zone material is radiated in turn by waves of $\lambda=350 \mathrm{~nm}$ and $540 \mathrm{~nm}$ respectively. The ratio of the stopping potential in the two cases is $2: 1$. The work function of the material is
1 $4.20 \mathrm{eV}$
2 $0.15 \mathrm{eV}$
3 $2.10 \mathrm{eV}$
4 $1.05 \mathrm{eV}$
Explanation:
D Given that, $\lambda_{1}=350 \mathrm{~nm}, \mathrm{~V}_{\mathrm{s}_{1}}=2 \mathrm{~V}_{\mathrm{s}_{2}}, \lambda_{2}=540 \mathrm{~nm}$ According to Einstein photoelectric equation - $\frac{\mathrm{hc}}{\lambda}=\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}$ $\lambda \propto \frac{1}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{1}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $\frac{540}{350}=\frac{\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $54\left(\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}\right)=35\left(\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}\right)$ $\mathrm{V}_{\mathrm{s}_{2}}=\frac{19}{16} \phi_{\mathrm{o}}$ From equation (i), we get - $\phi_{\mathrm{o}} =\frac{\mathrm{hc}}{\lambda_{1}}-\mathrm{V}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-2 \mathrm{~V}_{\mathrm{s}_{2}}$ $\phi_{\mathrm{o}} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{-9} \times 1.6 \times 10^{-19}}-2 \times \frac{19}{16} \phi_{\mathrm{o}}$ $\phi_{\mathrm{o}} =0.0105 \times 10^{2}$ $\phi_{\mathrm{o}} =1.05 \mathrm{eV}$
AIIMS-2000
Dual nature of radiation and Matter
142084
In a photo emissive cell with executing wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of fastest emitted electron will be-
1 greater than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
2 $\mathrm{v}\left(\frac{3}{4}\right)^{1 / 2}$
3 $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
4 less than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
Explanation:
A From Einstein photoelectric equation $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-\lambda\right)}{\lambda \lambda_{0}}}$ If wavelength is changed to $\left(\frac{3 \lambda}{4}\right)$ $\frac{1}{2} \mathrm{mv}^{\prime 2}=\frac{4 h \mathrm{c}}{3 \lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\lambda_{0}(3 \lambda / 4)}}$ Form equation (i) and (ii), we get- $\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\sqrt{\frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\frac{3}{4} \lambda \lambda_{0}} \times \frac{\lambda \lambda_{0}}{\lambda_{0}-\lambda}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{4}{3} \times \mathrm{v} \times \sqrt{\frac{\lambda_{0}-3 \lambda / 4}{\lambda_{0}-\lambda}}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3} \Rightarrow \mathrm{v}^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \mathrm{v}$
BCECE-2015
Dual nature of radiation and Matter
142086
A radio transmitter operates at a frequency of $880 \mathrm{kHz}$ and power of $10 \mathrm{~kW}$. The number of photons emitted per second is-
1 $13.27 \times 10^{4}$
2 $13.27 \times 10^{34}$
3 $1327 \times 10^{34}$
4 $1.71 \times 10^{31}$
Explanation:
D Given that, Frequency $(v)=880 \mathrm{kHz}$ Power $(\mathrm{P})=10 \mathrm{~kW}$ Number of photons emitted per second, $\mathrm{h}=\frac{\text { Power }}{\text { Energy of photon }}$ $\mathrm{h}=\frac{\text { Power }}{\mathrm{h} v}$ $\mathrm{~h}=\frac{10}{6.6 \times 10^{-34} \times 880}$ $\mathrm{~h}=1.72 \times 10^{31}$
BCECE-2011
Dual nature of radiation and Matter
142080
Assertion: Photoelectric saturation current increases with the increase in frequency of incident light. Reason: Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
D Photoelectric saturation current is independent of frequency. It only depends on the intensity of light.
AIIMS-2015
Dual nature of radiation and Matter
142087
Light of wavelength $5000 \AA$ falling on a sensitive surface. If the surface has received $10^{-}$ ${ }^{7} \mathbf{J}$ of energy, then the number of photons falling on the surface will be-
1 $5 \times 10^{11}$
2 $2.5 \times 10^{11}$
3 $3 \times 10^{11}$
4 none of these
Explanation:
B Given that, $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$ $\mathrm{E}=10^{-7} \mathrm{~J}$ $\therefore$ Number of photons $(n)=\frac{E}{h c / \lambda}=\frac{E \lambda}{h c}$ $\mathrm{n}=\frac{10^{-7} \times 5 \times 10^{7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=252 \times 10^{9}$ $\mathrm{n} \approx 2.5 \times 10^{11}$
142083
The surface of zone material is radiated in turn by waves of $\lambda=350 \mathrm{~nm}$ and $540 \mathrm{~nm}$ respectively. The ratio of the stopping potential in the two cases is $2: 1$. The work function of the material is
1 $4.20 \mathrm{eV}$
2 $0.15 \mathrm{eV}$
3 $2.10 \mathrm{eV}$
4 $1.05 \mathrm{eV}$
Explanation:
D Given that, $\lambda_{1}=350 \mathrm{~nm}, \mathrm{~V}_{\mathrm{s}_{1}}=2 \mathrm{~V}_{\mathrm{s}_{2}}, \lambda_{2}=540 \mathrm{~nm}$ According to Einstein photoelectric equation - $\frac{\mathrm{hc}}{\lambda}=\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}$ $\lambda \propto \frac{1}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}}}$ $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{1}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $\frac{540}{350}=\frac{\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}}{\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}}$ $54\left(\phi_{\mathrm{o}}+\mathrm{V}_{\mathrm{s}_{2}}\right)=35\left(\phi_{\mathrm{o}}+2 \mathrm{~V}_{\mathrm{s}_{2}}\right)$ $\mathrm{V}_{\mathrm{s}_{2}}=\frac{19}{16} \phi_{\mathrm{o}}$ From equation (i), we get - $\phi_{\mathrm{o}} =\frac{\mathrm{hc}}{\lambda_{1}}-\mathrm{V}_{\mathrm{s}_{1}}=\frac{\mathrm{hc}}{\lambda_{1}}-2 \mathrm{~V}_{\mathrm{s}_{2}}$ $\phi_{\mathrm{o}} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{-9} \times 1.6 \times 10^{-19}}-2 \times \frac{19}{16} \phi_{\mathrm{o}}$ $\phi_{\mathrm{o}} =0.0105 \times 10^{2}$ $\phi_{\mathrm{o}} =1.05 \mathrm{eV}$
AIIMS-2000
Dual nature of radiation and Matter
142084
In a photo emissive cell with executing wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of fastest emitted electron will be-
1 greater than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
2 $\mathrm{v}\left(\frac{3}{4}\right)^{1 / 2}$
3 $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
4 less than $\mathrm{v}\left(\frac{4}{3}\right)^{1 / 2}$
Explanation:
A From Einstein photoelectric equation $\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-\lambda\right)}{\lambda \lambda_{0}}}$ If wavelength is changed to $\left(\frac{3 \lambda}{4}\right)$ $\frac{1}{2} \mathrm{mv}^{\prime 2}=\frac{4 h \mathrm{c}}{3 \lambda}-\phi_{\mathrm{o}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{2 \mathrm{hc}}{\mathrm{m}} \frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\lambda_{0}(3 \lambda / 4)}}$ Form equation (i) and (ii), we get- $\frac{\mathrm{v}^{\prime}}{\mathrm{v}}=\sqrt{\frac{\left(\lambda_{0}-3 \lambda / 4\right)}{\frac{3}{4} \lambda \lambda_{0}} \times \frac{\lambda \lambda_{0}}{\lambda_{0}-\lambda}}$ $\mathrm{v}^{\prime}=\sqrt{\frac{4}{3} \times \mathrm{v} \times \sqrt{\frac{\lambda_{0}-3 \lambda / 4}{\lambda_{0}-\lambda}}}$ $\frac{\mathrm{v}^{\prime 2}}{\mathrm{v}^{2}}>\frac{4}{3} \Rightarrow \mathrm{v}^{\prime}>\left(\frac{4}{3}\right)^{1 / 2} \mathrm{v}$
BCECE-2015
Dual nature of radiation and Matter
142086
A radio transmitter operates at a frequency of $880 \mathrm{kHz}$ and power of $10 \mathrm{~kW}$. The number of photons emitted per second is-
1 $13.27 \times 10^{4}$
2 $13.27 \times 10^{34}$
3 $1327 \times 10^{34}$
4 $1.71 \times 10^{31}$
Explanation:
D Given that, Frequency $(v)=880 \mathrm{kHz}$ Power $(\mathrm{P})=10 \mathrm{~kW}$ Number of photons emitted per second, $\mathrm{h}=\frac{\text { Power }}{\text { Energy of photon }}$ $\mathrm{h}=\frac{\text { Power }}{\mathrm{h} v}$ $\mathrm{~h}=\frac{10}{6.6 \times 10^{-34} \times 880}$ $\mathrm{~h}=1.72 \times 10^{31}$
BCECE-2011
Dual nature of radiation and Matter
142080
Assertion: Photoelectric saturation current increases with the increase in frequency of incident light. Reason: Energy of incident photons increases with increase in frequency and as a result photoelectric current increases.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
D Photoelectric saturation current is independent of frequency. It only depends on the intensity of light.
AIIMS-2015
Dual nature of radiation and Matter
142087
Light of wavelength $5000 \AA$ falling on a sensitive surface. If the surface has received $10^{-}$ ${ }^{7} \mathbf{J}$ of energy, then the number of photons falling on the surface will be-
1 $5 \times 10^{11}$
2 $2.5 \times 10^{11}$
3 $3 \times 10^{11}$
4 none of these
Explanation:
B Given that, $\lambda=5000 \AA=5 \times 10^{-7} \mathrm{~m}$ $\mathrm{E}=10^{-7} \mathrm{~J}$ $\therefore$ Number of photons $(n)=\frac{E}{h c / \lambda}=\frac{E \lambda}{h c}$ $\mathrm{n}=\frac{10^{-7} \times 5 \times 10^{7}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}=252 \times 10^{9}$ $\mathrm{n} \approx 2.5 \times 10^{11}$
