142264
When silver is irradiated by ultraviolet light of $1000 \AA$, potential of $7.7 \mathrm{~V}$ is required to stop the photo electrons. The work function of silver will be:
1 $3.72 \mathrm{eV}$
2 $6.72 \mathrm{eV}$
3 $5.72 \mathrm{eV}$
4 $4.67 \mathrm{eV}$
Explanation:
D Given that, $\lambda=1000 \AA=10^{-7} \mathrm{~m}$ $\mathrm{~V}_{0}=7.7 \mathrm{~V}$ $\text { We know that, }$ $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=\frac{\mathrm{hc}}{\lambda}-\phi$ $\mathrm{E}_{\mathrm{K}}=\mathrm{eV} \mathrm{V}_{0}$ $\text { Work function }(\phi)=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}$ $\phi=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-7}}-1.6 \times 10^{-19} \times 7.7$ $\phi=19.8 \times 10^{-19}-12.32 \times 10^{-19}$ $\phi=7.48 \times 10^{-19} \mathrm{~J}$ $\phi=\frac{7.48 \times 10^{-19}}{1.6 \times 10^{-19}}=4.67 \mathrm{eV}$
JCECE-2004
Dual nature of radiation and Matter
142265
The energy of $\mathrm{eV}$ of red light of wavelength $\lambda=$ $6560 \AA$ is :
142266
Threshold wave length for lithium metal is 6250 . For photo emission. The wave length of the incident light must be
1 More than $6250 \square$
2 Exactly equal to 6250 m
3 Equal to or more than 6250 W
4 Equal to or less than 6250 ए
Explanation:
D Given that, $\lambda_{0}=6250 \square$ Photo electric effect equation- $\mathrm{h} v=h v_{0}+(\text { K.E. })_{\max }$ $\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{he}}{\lambda_{0}}+(\text { K.E. })_{\max }$ $v>v_{0}$ $\frac{\mathrm{c}}{\lambda}>\frac{\mathrm{c}}{\lambda_{0}}$ $\lambda_{0}>\lambda$ Thus, wavelength of the incident light must be equal to or less than $6250 \mathrm{C}$.
GUJCET 2014
Dual nature of radiation and Matter
142267
The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals $M$ and $N$. We may definitely conclude
1 Work function of $\mathrm{M}>$ work function of $\mathrm{N}$
2 Work function of $\mathrm{M} \lt $ work function of $\mathrm{N}$
3 Work function of $\mathrm{M}=$ work function of $\mathrm{N}$
4 At the threshold frequency of $\mathrm{M}$, the kinetic energy of the photoelectron emitted by $M$ is more than emitted by $\mathrm{N}$
Explanation:
A According to the photoelectric equation- $\mathrm{eV}_{0}=\mathrm{h} v-\phi_{0}$ $\phi=\mathrm{h} v_{0}$ Where, $\mathrm{h}=$ Planck's constant $v_{0}=$ Threshold frequency. As the threshold frequency of $\mathrm{M}$ is more thus that of $\mathrm{N}$, its work function is also more.
142264
When silver is irradiated by ultraviolet light of $1000 \AA$, potential of $7.7 \mathrm{~V}$ is required to stop the photo electrons. The work function of silver will be:
1 $3.72 \mathrm{eV}$
2 $6.72 \mathrm{eV}$
3 $5.72 \mathrm{eV}$
4 $4.67 \mathrm{eV}$
Explanation:
D Given that, $\lambda=1000 \AA=10^{-7} \mathrm{~m}$ $\mathrm{~V}_{0}=7.7 \mathrm{~V}$ $\text { We know that, }$ $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=\frac{\mathrm{hc}}{\lambda}-\phi$ $\mathrm{E}_{\mathrm{K}}=\mathrm{eV} \mathrm{V}_{0}$ $\text { Work function }(\phi)=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}$ $\phi=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-7}}-1.6 \times 10^{-19} \times 7.7$ $\phi=19.8 \times 10^{-19}-12.32 \times 10^{-19}$ $\phi=7.48 \times 10^{-19} \mathrm{~J}$ $\phi=\frac{7.48 \times 10^{-19}}{1.6 \times 10^{-19}}=4.67 \mathrm{eV}$
JCECE-2004
Dual nature of radiation and Matter
142265
The energy of $\mathrm{eV}$ of red light of wavelength $\lambda=$ $6560 \AA$ is :
142266
Threshold wave length for lithium metal is 6250 . For photo emission. The wave length of the incident light must be
1 More than $6250 \square$
2 Exactly equal to 6250 m
3 Equal to or more than 6250 W
4 Equal to or less than 6250 ए
Explanation:
D Given that, $\lambda_{0}=6250 \square$ Photo electric effect equation- $\mathrm{h} v=h v_{0}+(\text { K.E. })_{\max }$ $\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{he}}{\lambda_{0}}+(\text { K.E. })_{\max }$ $v>v_{0}$ $\frac{\mathrm{c}}{\lambda}>\frac{\mathrm{c}}{\lambda_{0}}$ $\lambda_{0}>\lambda$ Thus, wavelength of the incident light must be equal to or less than $6250 \mathrm{C}$.
GUJCET 2014
Dual nature of radiation and Matter
142267
The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals $M$ and $N$. We may definitely conclude
1 Work function of $\mathrm{M}>$ work function of $\mathrm{N}$
2 Work function of $\mathrm{M} \lt $ work function of $\mathrm{N}$
3 Work function of $\mathrm{M}=$ work function of $\mathrm{N}$
4 At the threshold frequency of $\mathrm{M}$, the kinetic energy of the photoelectron emitted by $M$ is more than emitted by $\mathrm{N}$
Explanation:
A According to the photoelectric equation- $\mathrm{eV}_{0}=\mathrm{h} v-\phi_{0}$ $\phi=\mathrm{h} v_{0}$ Where, $\mathrm{h}=$ Planck's constant $v_{0}=$ Threshold frequency. As the threshold frequency of $\mathrm{M}$ is more thus that of $\mathrm{N}$, its work function is also more.
142264
When silver is irradiated by ultraviolet light of $1000 \AA$, potential of $7.7 \mathrm{~V}$ is required to stop the photo electrons. The work function of silver will be:
1 $3.72 \mathrm{eV}$
2 $6.72 \mathrm{eV}$
3 $5.72 \mathrm{eV}$
4 $4.67 \mathrm{eV}$
Explanation:
D Given that, $\lambda=1000 \AA=10^{-7} \mathrm{~m}$ $\mathrm{~V}_{0}=7.7 \mathrm{~V}$ $\text { We know that, }$ $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=\frac{\mathrm{hc}}{\lambda}-\phi$ $\mathrm{E}_{\mathrm{K}}=\mathrm{eV} \mathrm{V}_{0}$ $\text { Work function }(\phi)=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}$ $\phi=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-7}}-1.6 \times 10^{-19} \times 7.7$ $\phi=19.8 \times 10^{-19}-12.32 \times 10^{-19}$ $\phi=7.48 \times 10^{-19} \mathrm{~J}$ $\phi=\frac{7.48 \times 10^{-19}}{1.6 \times 10^{-19}}=4.67 \mathrm{eV}$
JCECE-2004
Dual nature of radiation and Matter
142265
The energy of $\mathrm{eV}$ of red light of wavelength $\lambda=$ $6560 \AA$ is :
142266
Threshold wave length for lithium metal is 6250 . For photo emission. The wave length of the incident light must be
1 More than $6250 \square$
2 Exactly equal to 6250 m
3 Equal to or more than 6250 W
4 Equal to or less than 6250 ए
Explanation:
D Given that, $\lambda_{0}=6250 \square$ Photo electric effect equation- $\mathrm{h} v=h v_{0}+(\text { K.E. })_{\max }$ $\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{he}}{\lambda_{0}}+(\text { K.E. })_{\max }$ $v>v_{0}$ $\frac{\mathrm{c}}{\lambda}>\frac{\mathrm{c}}{\lambda_{0}}$ $\lambda_{0}>\lambda$ Thus, wavelength of the incident light must be equal to or less than $6250 \mathrm{C}$.
GUJCET 2014
Dual nature of radiation and Matter
142267
The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals $M$ and $N$. We may definitely conclude
1 Work function of $\mathrm{M}>$ work function of $\mathrm{N}$
2 Work function of $\mathrm{M} \lt $ work function of $\mathrm{N}$
3 Work function of $\mathrm{M}=$ work function of $\mathrm{N}$
4 At the threshold frequency of $\mathrm{M}$, the kinetic energy of the photoelectron emitted by $M$ is more than emitted by $\mathrm{N}$
Explanation:
A According to the photoelectric equation- $\mathrm{eV}_{0}=\mathrm{h} v-\phi_{0}$ $\phi=\mathrm{h} v_{0}$ Where, $\mathrm{h}=$ Planck's constant $v_{0}=$ Threshold frequency. As the threshold frequency of $\mathrm{M}$ is more thus that of $\mathrm{N}$, its work function is also more.
142264
When silver is irradiated by ultraviolet light of $1000 \AA$, potential of $7.7 \mathrm{~V}$ is required to stop the photo electrons. The work function of silver will be:
1 $3.72 \mathrm{eV}$
2 $6.72 \mathrm{eV}$
3 $5.72 \mathrm{eV}$
4 $4.67 \mathrm{eV}$
Explanation:
D Given that, $\lambda=1000 \AA=10^{-7} \mathrm{~m}$ $\mathrm{~V}_{0}=7.7 \mathrm{~V}$ $\text { We know that, }$ $\mathrm{E}_{\mathrm{K}}=\mathrm{h} v-\phi$ $\mathrm{E}_{\mathrm{K}}=\frac{\mathrm{hc}}{\lambda}-\phi$ $\mathrm{E}_{\mathrm{K}}=\mathrm{eV} \mathrm{V}_{0}$ $\text { Work function }(\phi)=\frac{\mathrm{hc}}{\lambda}-\mathrm{eV}$ $\phi=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{10^{-7}}-1.6 \times 10^{-19} \times 7.7$ $\phi=19.8 \times 10^{-19}-12.32 \times 10^{-19}$ $\phi=7.48 \times 10^{-19} \mathrm{~J}$ $\phi=\frac{7.48 \times 10^{-19}}{1.6 \times 10^{-19}}=4.67 \mathrm{eV}$
JCECE-2004
Dual nature of radiation and Matter
142265
The energy of $\mathrm{eV}$ of red light of wavelength $\lambda=$ $6560 \AA$ is :
142266
Threshold wave length for lithium metal is 6250 . For photo emission. The wave length of the incident light must be
1 More than $6250 \square$
2 Exactly equal to 6250 m
3 Equal to or more than 6250 W
4 Equal to or less than 6250 ए
Explanation:
D Given that, $\lambda_{0}=6250 \square$ Photo electric effect equation- $\mathrm{h} v=h v_{0}+(\text { K.E. })_{\max }$ $\frac{\mathrm{hc}}{\lambda}=\frac{\mathrm{he}}{\lambda_{0}}+(\text { K.E. })_{\max }$ $v>v_{0}$ $\frac{\mathrm{c}}{\lambda}>\frac{\mathrm{c}}{\lambda_{0}}$ $\lambda_{0}>\lambda$ Thus, wavelength of the incident light must be equal to or less than $6250 \mathrm{C}$.
GUJCET 2014
Dual nature of radiation and Matter
142267
The graph of kinetic energy of photoelectron versus frequency of the incident radiation is shown for two metals $M$ and $N$. We may definitely conclude
1 Work function of $\mathrm{M}>$ work function of $\mathrm{N}$
2 Work function of $\mathrm{M} \lt $ work function of $\mathrm{N}$
3 Work function of $\mathrm{M}=$ work function of $\mathrm{N}$
4 At the threshold frequency of $\mathrm{M}$, the kinetic energy of the photoelectron emitted by $M$ is more than emitted by $\mathrm{N}$
Explanation:
A According to the photoelectric equation- $\mathrm{eV}_{0}=\mathrm{h} v-\phi_{0}$ $\phi=\mathrm{h} v_{0}$ Where, $\mathrm{h}=$ Planck's constant $v_{0}=$ Threshold frequency. As the threshold frequency of $\mathrm{M}$ is more thus that of $\mathrm{N}$, its work function is also more.
