142088
The threshold frequency for certain metal is $\mathbf{3 . 3}$ $\times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, the cut-off voltage of the photoelectric current will be -
1 $4.9 \mathrm{~V}$
2 $3.0 \mathrm{~V}$
3 $2.0 \mathrm{~V}$
4 $1.0 \mathrm{~V}$
Explanation:
C Given that, $v_{\mathrm{o}}=3.3 \times 10^{14} \mathrm{~Hz}$ $v=8.2 \times 10^{14} \mathrm{~Hz}$ We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ $\mathrm{V}_{\mathrm{o}}=\frac{\mathrm{h}\left(v-\mathrm{v}_{\mathrm{o}}\right)}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34}(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}$ $\mathrm{~V}_{\mathrm{o}}=2.029$ $\mathrm{~V}_{\mathrm{o}} \approx 2.0 \mathrm{~V}$
BCECE-2006
Dual nature of radiation and Matter
142089
Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy $E$. If $E$ is to be increased to exactly twice its original value, the wavelength changes to $\lambda^{\prime}$, where-
1 $\lambda^{\prime}$ is less than $\frac{\lambda}{2}$
2 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$
3 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$ but less than $\lambda$
4 $\lambda^{\prime}$ is exactly equal to $\frac{\lambda}{2}$
Explanation:
C According to Einstein photoelectric equation $\frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=2 \mathrm{E}+\phi_{\mathrm{o}}$ From equation (i) and (ii), we get- $\frac{\lambda^{\prime}}{\lambda}=\left(\frac{\mathrm{E}+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $\frac{\lambda^{\prime}}{\lambda} \lt 1$ $\frac{\lambda^{\prime}}{\lambda} \lt \lambda$ Also $\frac{\lambda^{\prime}}{\lambda} =\left(\frac{E+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $=\frac{1}{2}\left[\frac{\mathrm{E}+\phi_{\mathrm{o}}}{\mathrm{E}+\phi_{\mathrm{o}} / 2}\right]$ $\frac{\lambda^{\prime}}{\lambda} >\frac{1}{2}$ So, from (iii) and (iv) $\lambda>\lambda^{\prime}>\frac{\lambda}{2}$
BCECE-2010
Dual nature of radiation and Matter
142092
The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$. The corresponding change in stopping potential is
142093
Light of wavelength ' $\lambda$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will
1 increase
2 decrease
3 be zero
4 become exactly half
Explanation:
A Incident photon (E) $=\mathrm{h} v$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E} \propto \frac{1}{\lambda}$ If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect, $\frac{\mathrm{hc}}{\lambda}=\mathrm{KE}_{\max }+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{eV}_{\mathrm{o}}+\phi_{\mathrm{o}}$ $\mathrm{eV}_{\mathrm{o}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right]$ $\mathrm{V}_{\mathrm{o}} \propto\left(\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right)$ Hence, if wavelength decreases, stopping potential increases.
142088
The threshold frequency for certain metal is $\mathbf{3 . 3}$ $\times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, the cut-off voltage of the photoelectric current will be -
1 $4.9 \mathrm{~V}$
2 $3.0 \mathrm{~V}$
3 $2.0 \mathrm{~V}$
4 $1.0 \mathrm{~V}$
Explanation:
C Given that, $v_{\mathrm{o}}=3.3 \times 10^{14} \mathrm{~Hz}$ $v=8.2 \times 10^{14} \mathrm{~Hz}$ We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ $\mathrm{V}_{\mathrm{o}}=\frac{\mathrm{h}\left(v-\mathrm{v}_{\mathrm{o}}\right)}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34}(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}$ $\mathrm{~V}_{\mathrm{o}}=2.029$ $\mathrm{~V}_{\mathrm{o}} \approx 2.0 \mathrm{~V}$
BCECE-2006
Dual nature of radiation and Matter
142089
Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy $E$. If $E$ is to be increased to exactly twice its original value, the wavelength changes to $\lambda^{\prime}$, where-
1 $\lambda^{\prime}$ is less than $\frac{\lambda}{2}$
2 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$
3 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$ but less than $\lambda$
4 $\lambda^{\prime}$ is exactly equal to $\frac{\lambda}{2}$
Explanation:
C According to Einstein photoelectric equation $\frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=2 \mathrm{E}+\phi_{\mathrm{o}}$ From equation (i) and (ii), we get- $\frac{\lambda^{\prime}}{\lambda}=\left(\frac{\mathrm{E}+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $\frac{\lambda^{\prime}}{\lambda} \lt 1$ $\frac{\lambda^{\prime}}{\lambda} \lt \lambda$ Also $\frac{\lambda^{\prime}}{\lambda} =\left(\frac{E+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $=\frac{1}{2}\left[\frac{\mathrm{E}+\phi_{\mathrm{o}}}{\mathrm{E}+\phi_{\mathrm{o}} / 2}\right]$ $\frac{\lambda^{\prime}}{\lambda} >\frac{1}{2}$ So, from (iii) and (iv) $\lambda>\lambda^{\prime}>\frac{\lambda}{2}$
BCECE-2010
Dual nature of radiation and Matter
142092
The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$. The corresponding change in stopping potential is
142093
Light of wavelength ' $\lambda$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will
1 increase
2 decrease
3 be zero
4 become exactly half
Explanation:
A Incident photon (E) $=\mathrm{h} v$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E} \propto \frac{1}{\lambda}$ If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect, $\frac{\mathrm{hc}}{\lambda}=\mathrm{KE}_{\max }+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{eV}_{\mathrm{o}}+\phi_{\mathrm{o}}$ $\mathrm{eV}_{\mathrm{o}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right]$ $\mathrm{V}_{\mathrm{o}} \propto\left(\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right)$ Hence, if wavelength decreases, stopping potential increases.
142088
The threshold frequency for certain metal is $\mathbf{3 . 3}$ $\times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, the cut-off voltage of the photoelectric current will be -
1 $4.9 \mathrm{~V}$
2 $3.0 \mathrm{~V}$
3 $2.0 \mathrm{~V}$
4 $1.0 \mathrm{~V}$
Explanation:
C Given that, $v_{\mathrm{o}}=3.3 \times 10^{14} \mathrm{~Hz}$ $v=8.2 \times 10^{14} \mathrm{~Hz}$ We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ $\mathrm{V}_{\mathrm{o}}=\frac{\mathrm{h}\left(v-\mathrm{v}_{\mathrm{o}}\right)}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34}(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}$ $\mathrm{~V}_{\mathrm{o}}=2.029$ $\mathrm{~V}_{\mathrm{o}} \approx 2.0 \mathrm{~V}$
BCECE-2006
Dual nature of radiation and Matter
142089
Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy $E$. If $E$ is to be increased to exactly twice its original value, the wavelength changes to $\lambda^{\prime}$, where-
1 $\lambda^{\prime}$ is less than $\frac{\lambda}{2}$
2 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$
3 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$ but less than $\lambda$
4 $\lambda^{\prime}$ is exactly equal to $\frac{\lambda}{2}$
Explanation:
C According to Einstein photoelectric equation $\frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=2 \mathrm{E}+\phi_{\mathrm{o}}$ From equation (i) and (ii), we get- $\frac{\lambda^{\prime}}{\lambda}=\left(\frac{\mathrm{E}+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $\frac{\lambda^{\prime}}{\lambda} \lt 1$ $\frac{\lambda^{\prime}}{\lambda} \lt \lambda$ Also $\frac{\lambda^{\prime}}{\lambda} =\left(\frac{E+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $=\frac{1}{2}\left[\frac{\mathrm{E}+\phi_{\mathrm{o}}}{\mathrm{E}+\phi_{\mathrm{o}} / 2}\right]$ $\frac{\lambda^{\prime}}{\lambda} >\frac{1}{2}$ So, from (iii) and (iv) $\lambda>\lambda^{\prime}>\frac{\lambda}{2}$
BCECE-2010
Dual nature of radiation and Matter
142092
The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$. The corresponding change in stopping potential is
142093
Light of wavelength ' $\lambda$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will
1 increase
2 decrease
3 be zero
4 become exactly half
Explanation:
A Incident photon (E) $=\mathrm{h} v$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E} \propto \frac{1}{\lambda}$ If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect, $\frac{\mathrm{hc}}{\lambda}=\mathrm{KE}_{\max }+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{eV}_{\mathrm{o}}+\phi_{\mathrm{o}}$ $\mathrm{eV}_{\mathrm{o}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right]$ $\mathrm{V}_{\mathrm{o}} \propto\left(\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right)$ Hence, if wavelength decreases, stopping potential increases.
142088
The threshold frequency for certain metal is $\mathbf{3 . 3}$ $\times 10^{14} \mathrm{~Hz}$. If light of frequency $8.2 \times 10^{14} \mathrm{~Hz}$ is incident on the metal, the cut-off voltage of the photoelectric current will be -
1 $4.9 \mathrm{~V}$
2 $3.0 \mathrm{~V}$
3 $2.0 \mathrm{~V}$
4 $1.0 \mathrm{~V}$
Explanation:
C Given that, $v_{\mathrm{o}}=3.3 \times 10^{14} \mathrm{~Hz}$ $v=8.2 \times 10^{14} \mathrm{~Hz}$ We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ $\mathrm{V}_{\mathrm{o}}=\frac{\mathrm{h}\left(v-\mathrm{v}_{\mathrm{o}}\right)}{\mathrm{e}}$ $\mathrm{V}_{\mathrm{o}}=\frac{6.62 \times 10^{-34}(8.2-3.3) \times 10^{14}}{1.6 \times 10^{-19}}$ $\mathrm{~V}_{\mathrm{o}}=2.029$ $\mathrm{~V}_{\mathrm{o}} \approx 2.0 \mathrm{~V}$
BCECE-2006
Dual nature of radiation and Matter
142089
Light of wavelength $\lambda$, strikes a photoelectric surface and electrons are ejected with an energy $E$. If $E$ is to be increased to exactly twice its original value, the wavelength changes to $\lambda^{\prime}$, where-
1 $\lambda^{\prime}$ is less than $\frac{\lambda}{2}$
2 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$
3 $\lambda^{\prime}$ is greater than $\frac{\lambda}{2}$ but less than $\lambda$
4 $\lambda^{\prime}$ is exactly equal to $\frac{\lambda}{2}$
Explanation:
C According to Einstein photoelectric equation $\frac{\mathrm{hc}}{\lambda}=\mathrm{E}+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=2 \mathrm{E}+\phi_{\mathrm{o}}$ From equation (i) and (ii), we get- $\frac{\lambda^{\prime}}{\lambda}=\left(\frac{\mathrm{E}+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $\frac{\lambda^{\prime}}{\lambda} \lt 1$ $\frac{\lambda^{\prime}}{\lambda} \lt \lambda$ Also $\frac{\lambda^{\prime}}{\lambda} =\left(\frac{E+\phi_{\mathrm{o}}}{2 \mathrm{E}+\phi_{\mathrm{o}}}\right)$ $=\frac{1}{2}\left[\frac{\mathrm{E}+\phi_{\mathrm{o}}}{\mathrm{E}+\phi_{\mathrm{o}} / 2}\right]$ $\frac{\lambda^{\prime}}{\lambda} >\frac{1}{2}$ So, from (iii) and (iv) $\lambda>\lambda^{\prime}>\frac{\lambda}{2}$
BCECE-2010
Dual nature of radiation and Matter
142092
The wavelength of incident light falling on a photosensitive surface is changed from $2000 \AA$ to $2100 \AA$. The corresponding change in stopping potential is
142093
Light of wavelength ' $\lambda$ ' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with same velocity then stopping potential will
1 increase
2 decrease
3 be zero
4 become exactly half
Explanation:
A Incident photon (E) $=\mathrm{h} v$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E} \propto \frac{1}{\lambda}$ If the incident wavelength is decreased then energy of the incident light is increased from Einstein photoelectric effect, $\frac{\mathrm{hc}}{\lambda}=\mathrm{KE}_{\max }+\phi_{\mathrm{o}}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{eV}_{\mathrm{o}}+\phi_{\mathrm{o}}$ $\mathrm{eV}_{\mathrm{o}}=\mathrm{hc}\left[\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right]$ $\mathrm{V}_{\mathrm{o}} \propto\left(\frac{1}{\lambda}-\frac{1}{\lambda_{\mathrm{o}}}\right)$ Hence, if wavelength decreases, stopping potential increases.
