142297
Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is
1 red
2 blue
3 green
4 violet
Explanation:
D We know that, $\mathrm{v}=\mathrm{f} \lambda \Rightarrow \mathrm{f} \propto \frac{1}{\lambda}$ Then, frequency is invesse to wavelength for constant velocity. Order of wavelength of given coloured- $\lambda_{\mathrm{v}} \lt \lambda_{\mathrm{b}} \lt \lambda_{\mathrm{g}} \lt \lambda_{\mathrm{r}}$ Frequency order- $f_{v}>f_{b}>f_{g}>f_{r}$ Energy order- $\mathrm{E}_{\mathrm{v}}>\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{g}}>\mathrm{E}_{\mathrm{r}}$ Hence, violet light can produce photo-electrons.
J and K CET-2015
Dual nature of radiation and Matter
142299
The energy of photon of light is $3 \mathrm{ev}$. Then, the wavelength of photon must be
1 $4125 \mathrm{~nm}$
2 $412.5 \mathrm{~nm}$
3 $41.250 \mathrm{~nm}$
4 $4 \mathrm{~nm}$
Explanation:
B Given that, Energy of photon of light $(\mathrm{E})=3 \mathrm{eV}$ We know that, Photo Energy $(E)=h v=3 e V$ $v=\mathrm{c} / \lambda$ Where, $\mathrm{h}=$ Plank's constat $=6.63 \times 10^{-34} \mathrm{JS}$ $\mathrm{c}=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ Form equation (i) and (ii), we get- $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{-19}}$ $\lambda=4.143 \times 10^{-7} \mathrm{~m}$ $\lambda=414.3 \times 10^{-9} \mathrm{~m}$ $\lambda=414.3 \mathrm{~nm}$ $\lambda \approx 412.5 \mathrm{~nm}$
UP CPMT-2009
Dual nature of radiation and Matter
142300
If work function of a metal is $4.2 \mathrm{eV}$, the cut-of wavelength is
1 $8000 \AA$
2 $7000 \AA$
3 $1472 \AA$
4 $2950 \AA$
Explanation:
D Given that, work function $(\phi)=4.2 \mathrm{eV}=4.2$ $\times 1.6 \times 10^{-19}=6.72 \times 10^{-19} \mathrm{~J}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{Js}$ We know that, $\lambda_{0}=\frac{\mathrm{hc}}{\phi}$ $\lambda_{0}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6.72 \times 10^{-19}}=2.9464 \times 10^{-7} \mathrm{~m}$ $\lambda_{0} \sqcup 2950 \AA$
UP CPMT-2004
Dual nature of radiation and Matter
142301
A light of wavelength $5000 \AA$ falls on a sensitive plate with photoelectric work function $1.90 \mathrm{eV}$. Kinetic energy of the emitted photoelectrons will be (Given, $h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
142297
Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is
1 red
2 blue
3 green
4 violet
Explanation:
D We know that, $\mathrm{v}=\mathrm{f} \lambda \Rightarrow \mathrm{f} \propto \frac{1}{\lambda}$ Then, frequency is invesse to wavelength for constant velocity. Order of wavelength of given coloured- $\lambda_{\mathrm{v}} \lt \lambda_{\mathrm{b}} \lt \lambda_{\mathrm{g}} \lt \lambda_{\mathrm{r}}$ Frequency order- $f_{v}>f_{b}>f_{g}>f_{r}$ Energy order- $\mathrm{E}_{\mathrm{v}}>\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{g}}>\mathrm{E}_{\mathrm{r}}$ Hence, violet light can produce photo-electrons.
J and K CET-2015
Dual nature of radiation and Matter
142299
The energy of photon of light is $3 \mathrm{ev}$. Then, the wavelength of photon must be
1 $4125 \mathrm{~nm}$
2 $412.5 \mathrm{~nm}$
3 $41.250 \mathrm{~nm}$
4 $4 \mathrm{~nm}$
Explanation:
B Given that, Energy of photon of light $(\mathrm{E})=3 \mathrm{eV}$ We know that, Photo Energy $(E)=h v=3 e V$ $v=\mathrm{c} / \lambda$ Where, $\mathrm{h}=$ Plank's constat $=6.63 \times 10^{-34} \mathrm{JS}$ $\mathrm{c}=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ Form equation (i) and (ii), we get- $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{-19}}$ $\lambda=4.143 \times 10^{-7} \mathrm{~m}$ $\lambda=414.3 \times 10^{-9} \mathrm{~m}$ $\lambda=414.3 \mathrm{~nm}$ $\lambda \approx 412.5 \mathrm{~nm}$
UP CPMT-2009
Dual nature of radiation and Matter
142300
If work function of a metal is $4.2 \mathrm{eV}$, the cut-of wavelength is
1 $8000 \AA$
2 $7000 \AA$
3 $1472 \AA$
4 $2950 \AA$
Explanation:
D Given that, work function $(\phi)=4.2 \mathrm{eV}=4.2$ $\times 1.6 \times 10^{-19}=6.72 \times 10^{-19} \mathrm{~J}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{Js}$ We know that, $\lambda_{0}=\frac{\mathrm{hc}}{\phi}$ $\lambda_{0}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6.72 \times 10^{-19}}=2.9464 \times 10^{-7} \mathrm{~m}$ $\lambda_{0} \sqcup 2950 \AA$
UP CPMT-2004
Dual nature of radiation and Matter
142301
A light of wavelength $5000 \AA$ falls on a sensitive plate with photoelectric work function $1.90 \mathrm{eV}$. Kinetic energy of the emitted photoelectrons will be (Given, $h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
142297
Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is
1 red
2 blue
3 green
4 violet
Explanation:
D We know that, $\mathrm{v}=\mathrm{f} \lambda \Rightarrow \mathrm{f} \propto \frac{1}{\lambda}$ Then, frequency is invesse to wavelength for constant velocity. Order of wavelength of given coloured- $\lambda_{\mathrm{v}} \lt \lambda_{\mathrm{b}} \lt \lambda_{\mathrm{g}} \lt \lambda_{\mathrm{r}}$ Frequency order- $f_{v}>f_{b}>f_{g}>f_{r}$ Energy order- $\mathrm{E}_{\mathrm{v}}>\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{g}}>\mathrm{E}_{\mathrm{r}}$ Hence, violet light can produce photo-electrons.
J and K CET-2015
Dual nature of radiation and Matter
142299
The energy of photon of light is $3 \mathrm{ev}$. Then, the wavelength of photon must be
1 $4125 \mathrm{~nm}$
2 $412.5 \mathrm{~nm}$
3 $41.250 \mathrm{~nm}$
4 $4 \mathrm{~nm}$
Explanation:
B Given that, Energy of photon of light $(\mathrm{E})=3 \mathrm{eV}$ We know that, Photo Energy $(E)=h v=3 e V$ $v=\mathrm{c} / \lambda$ Where, $\mathrm{h}=$ Plank's constat $=6.63 \times 10^{-34} \mathrm{JS}$ $\mathrm{c}=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ Form equation (i) and (ii), we get- $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{-19}}$ $\lambda=4.143 \times 10^{-7} \mathrm{~m}$ $\lambda=414.3 \times 10^{-9} \mathrm{~m}$ $\lambda=414.3 \mathrm{~nm}$ $\lambda \approx 412.5 \mathrm{~nm}$
UP CPMT-2009
Dual nature of radiation and Matter
142300
If work function of a metal is $4.2 \mathrm{eV}$, the cut-of wavelength is
1 $8000 \AA$
2 $7000 \AA$
3 $1472 \AA$
4 $2950 \AA$
Explanation:
D Given that, work function $(\phi)=4.2 \mathrm{eV}=4.2$ $\times 1.6 \times 10^{-19}=6.72 \times 10^{-19} \mathrm{~J}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{Js}$ We know that, $\lambda_{0}=\frac{\mathrm{hc}}{\phi}$ $\lambda_{0}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6.72 \times 10^{-19}}=2.9464 \times 10^{-7} \mathrm{~m}$ $\lambda_{0} \sqcup 2950 \AA$
UP CPMT-2004
Dual nature of radiation and Matter
142301
A light of wavelength $5000 \AA$ falls on a sensitive plate with photoelectric work function $1.90 \mathrm{eV}$. Kinetic energy of the emitted photoelectrons will be (Given, $h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
142297
Red, blue, green and violet colour lights are one by one made incident on a photocathode. It is observed that only one colour light produces photoelectrons. That light is
1 red
2 blue
3 green
4 violet
Explanation:
D We know that, $\mathrm{v}=\mathrm{f} \lambda \Rightarrow \mathrm{f} \propto \frac{1}{\lambda}$ Then, frequency is invesse to wavelength for constant velocity. Order of wavelength of given coloured- $\lambda_{\mathrm{v}} \lt \lambda_{\mathrm{b}} \lt \lambda_{\mathrm{g}} \lt \lambda_{\mathrm{r}}$ Frequency order- $f_{v}>f_{b}>f_{g}>f_{r}$ Energy order- $\mathrm{E}_{\mathrm{v}}>\mathrm{E}_{\mathrm{b}}>\mathrm{E}_{\mathrm{g}}>\mathrm{E}_{\mathrm{r}}$ Hence, violet light can produce photo-electrons.
J and K CET-2015
Dual nature of radiation and Matter
142299
The energy of photon of light is $3 \mathrm{ev}$. Then, the wavelength of photon must be
1 $4125 \mathrm{~nm}$
2 $412.5 \mathrm{~nm}$
3 $41.250 \mathrm{~nm}$
4 $4 \mathrm{~nm}$
Explanation:
B Given that, Energy of photon of light $(\mathrm{E})=3 \mathrm{eV}$ We know that, Photo Energy $(E)=h v=3 e V$ $v=\mathrm{c} / \lambda$ Where, $\mathrm{h}=$ Plank's constat $=6.63 \times 10^{-34} \mathrm{JS}$ $\mathrm{c}=$ Speed of light $=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}$ $1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}$ Form equation (i) and (ii), we get- $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{3 \times 1.6 \times 10^{-19}}$ $\lambda=4.143 \times 10^{-7} \mathrm{~m}$ $\lambda=414.3 \times 10^{-9} \mathrm{~m}$ $\lambda=414.3 \mathrm{~nm}$ $\lambda \approx 412.5 \mathrm{~nm}$
UP CPMT-2009
Dual nature of radiation and Matter
142300
If work function of a metal is $4.2 \mathrm{eV}$, the cut-of wavelength is
1 $8000 \AA$
2 $7000 \AA$
3 $1472 \AA$
4 $2950 \AA$
Explanation:
D Given that, work function $(\phi)=4.2 \mathrm{eV}=4.2$ $\times 1.6 \times 10^{-19}=6.72 \times 10^{-19} \mathrm{~J}$ Speed of light $(\mathrm{c})=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Planck's constant $(\mathrm{h})=6.6 \times 10^{-34} \mathrm{Js}$ We know that, $\lambda_{0}=\frac{\mathrm{hc}}{\phi}$ $\lambda_{0}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6.72 \times 10^{-19}}=2.9464 \times 10^{-7} \mathrm{~m}$ $\lambda_{0} \sqcup 2950 \AA$
UP CPMT-2004
Dual nature of radiation and Matter
142301
A light of wavelength $5000 \AA$ falls on a sensitive plate with photoelectric work function $1.90 \mathrm{eV}$. Kinetic energy of the emitted photoelectrons will be (Given, $h=6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ )
