142274
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$. Then the work function of the metal must be
1 $\frac{h v}{2}$
2 hv
3 $2 \mathrm{hv}$
4 none of the above
Explanation:
A We know that, $\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ Now, the frequency is increased to $\frac{3 \mathrm{v}}{2}$ $\frac{3 \mathrm{hv}}{2}=\phi+2 \mathrm{~V}_{\mathrm{s}}$ From equation (i) and (ii) we get - $\phi=\frac{\mathrm{hv}}{2}$ Work function $=\frac{\mathrm{hv}}{2}$
AIIMS-2009
Dual nature of radiation and Matter
142275
A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
A Given, $\phi=2 \mathrm{eV}, \mathrm{h} v=4 \mathrm{eV}$ According to Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{h} v-\phi$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{eV}-2 \mathrm{eV}$ Stopping potential- $\mathrm{eV}_{0}=2 \mathrm{eV}$ $\mathrm{V}_{0}=2 \mathrm{~V}$
AIIMS-2004
Dual nature of radiation and Matter
142277
Two identical metal plates show photoelectric effect by a light of wavelength $\lambda_{1}$ on plate 1 and $\lambda_{2}$ on plate 2 (where $\lambda_{1}=2 \lambda_{2}$ ). The maximum kinetic energy will be-
1 $2 \mathrm{~K}_{2}=\mathrm{K}_{1}$
2 $\mathrm{K}_{1} \lt \mathrm{K}_{2} / 2$
3 $\mathrm{K}_{1}>\mathrm{K}_{2} / 2$
4 $2 \mathrm{~K}_{1}=\mathrm{K}_{2}$
Explanation:
B Maximum kinetic energy, $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Condition I- $\because \quad \lambda_{1} =2 \lambda_{2}$ $\mathrm{~K}_{1} =\frac{\mathrm{hc}}{2 \lambda_{2}}-\phi_{\mathrm{o}}$ Condition II- $\because \quad \lambda=\lambda_{2}$ $\mathrm{~K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Dividing both side of the above equation by 2 , $\frac{\mathrm{K}_{2}}{2}=\frac{\mathrm{hc}}{2 \lambda_{2}}-\frac{\phi_{\mathrm{o}}}{2}$ From equation (i) and (ii), we get- $\frac{\mathrm{K}_{2}}{2}-\mathrm{K}_{1}=\frac{\phi_{\mathrm{o}}}{2}$ $\frac{\mathrm{K}_{2}}{2}=\mathrm{K}_{1}+\frac{\phi_{\mathrm{o}}}{2}$ So, from equation (iii) we can say, $\frac{\mathrm{K}_{2}}{2}>\mathrm{K}_{1}$
BCECE-2016
Dual nature of radiation and Matter
142278
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 \mathrm{~cm}^{2}$, then the total momentum delivered on the surface during a span of 30 min is-
A Given, $\mathrm{I}=18 \mathrm{~W} / \mathrm{cm}^{2}, \mathrm{~A}=20 \mathrm{~cm}^{2}, \Delta \mathrm{t}=30$ $\min =30 \times 60=1800 \mathrm{sec}$ Total energy falling on the surface $\mathrm{U}=\mathrm{IA} \Delta \mathrm{t}$ $\mathrm{U}=18 \times 20 \times 1800 \mathrm{~J}$ $\mathrm{U}=6.48 \times 10^{5} \mathrm{~J}$ Momentum, $\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}$ $\mathrm{p}=\frac{6.48 \times 10^{5}}{3 \times 10^{8}}$ $\mathrm{p}=2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
BCECE-2014
Dual nature of radiation and Matter
142279
No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 \mathrm{~nm}$. The work function of the metal is approximately equal to-
1 $3 \times 10^{-16} \mathrm{~J}$
2 $3 \times 10^{-19} \mathrm{~J}$
3 $3 \times 10^{-20} \mathrm{~J}$
4 $3 \times 10^{-22} \mathrm{~J}$
Explanation:
B Given that, $\lambda_{\mathrm{o}}=6000 \stackrel{\circ}{\mathrm{A}}=6 \times 10^{-7} \mathrm{~m}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ Since, we know that work function is given as- $\phi_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}}$ $\phi_{\mathrm{o}}=3 \times 10^{-19} \mathrm{~J}$
142274
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$. Then the work function of the metal must be
1 $\frac{h v}{2}$
2 hv
3 $2 \mathrm{hv}$
4 none of the above
Explanation:
A We know that, $\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ Now, the frequency is increased to $\frac{3 \mathrm{v}}{2}$ $\frac{3 \mathrm{hv}}{2}=\phi+2 \mathrm{~V}_{\mathrm{s}}$ From equation (i) and (ii) we get - $\phi=\frac{\mathrm{hv}}{2}$ Work function $=\frac{\mathrm{hv}}{2}$
AIIMS-2009
Dual nature of radiation and Matter
142275
A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
A Given, $\phi=2 \mathrm{eV}, \mathrm{h} v=4 \mathrm{eV}$ According to Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{h} v-\phi$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{eV}-2 \mathrm{eV}$ Stopping potential- $\mathrm{eV}_{0}=2 \mathrm{eV}$ $\mathrm{V}_{0}=2 \mathrm{~V}$
AIIMS-2004
Dual nature of radiation and Matter
142277
Two identical metal plates show photoelectric effect by a light of wavelength $\lambda_{1}$ on plate 1 and $\lambda_{2}$ on plate 2 (where $\lambda_{1}=2 \lambda_{2}$ ). The maximum kinetic energy will be-
1 $2 \mathrm{~K}_{2}=\mathrm{K}_{1}$
2 $\mathrm{K}_{1} \lt \mathrm{K}_{2} / 2$
3 $\mathrm{K}_{1}>\mathrm{K}_{2} / 2$
4 $2 \mathrm{~K}_{1}=\mathrm{K}_{2}$
Explanation:
B Maximum kinetic energy, $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Condition I- $\because \quad \lambda_{1} =2 \lambda_{2}$ $\mathrm{~K}_{1} =\frac{\mathrm{hc}}{2 \lambda_{2}}-\phi_{\mathrm{o}}$ Condition II- $\because \quad \lambda=\lambda_{2}$ $\mathrm{~K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Dividing both side of the above equation by 2 , $\frac{\mathrm{K}_{2}}{2}=\frac{\mathrm{hc}}{2 \lambda_{2}}-\frac{\phi_{\mathrm{o}}}{2}$ From equation (i) and (ii), we get- $\frac{\mathrm{K}_{2}}{2}-\mathrm{K}_{1}=\frac{\phi_{\mathrm{o}}}{2}$ $\frac{\mathrm{K}_{2}}{2}=\mathrm{K}_{1}+\frac{\phi_{\mathrm{o}}}{2}$ So, from equation (iii) we can say, $\frac{\mathrm{K}_{2}}{2}>\mathrm{K}_{1}$
BCECE-2016
Dual nature of radiation and Matter
142278
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 \mathrm{~cm}^{2}$, then the total momentum delivered on the surface during a span of 30 min is-
A Given, $\mathrm{I}=18 \mathrm{~W} / \mathrm{cm}^{2}, \mathrm{~A}=20 \mathrm{~cm}^{2}, \Delta \mathrm{t}=30$ $\min =30 \times 60=1800 \mathrm{sec}$ Total energy falling on the surface $\mathrm{U}=\mathrm{IA} \Delta \mathrm{t}$ $\mathrm{U}=18 \times 20 \times 1800 \mathrm{~J}$ $\mathrm{U}=6.48 \times 10^{5} \mathrm{~J}$ Momentum, $\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}$ $\mathrm{p}=\frac{6.48 \times 10^{5}}{3 \times 10^{8}}$ $\mathrm{p}=2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
BCECE-2014
Dual nature of radiation and Matter
142279
No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 \mathrm{~nm}$. The work function of the metal is approximately equal to-
1 $3 \times 10^{-16} \mathrm{~J}$
2 $3 \times 10^{-19} \mathrm{~J}$
3 $3 \times 10^{-20} \mathrm{~J}$
4 $3 \times 10^{-22} \mathrm{~J}$
Explanation:
B Given that, $\lambda_{\mathrm{o}}=6000 \stackrel{\circ}{\mathrm{A}}=6 \times 10^{-7} \mathrm{~m}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ Since, we know that work function is given as- $\phi_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}}$ $\phi_{\mathrm{o}}=3 \times 10^{-19} \mathrm{~J}$
142274
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$. Then the work function of the metal must be
1 $\frac{h v}{2}$
2 hv
3 $2 \mathrm{hv}$
4 none of the above
Explanation:
A We know that, $\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ Now, the frequency is increased to $\frac{3 \mathrm{v}}{2}$ $\frac{3 \mathrm{hv}}{2}=\phi+2 \mathrm{~V}_{\mathrm{s}}$ From equation (i) and (ii) we get - $\phi=\frac{\mathrm{hv}}{2}$ Work function $=\frac{\mathrm{hv}}{2}$
AIIMS-2009
Dual nature of radiation and Matter
142275
A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
A Given, $\phi=2 \mathrm{eV}, \mathrm{h} v=4 \mathrm{eV}$ According to Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{h} v-\phi$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{eV}-2 \mathrm{eV}$ Stopping potential- $\mathrm{eV}_{0}=2 \mathrm{eV}$ $\mathrm{V}_{0}=2 \mathrm{~V}$
AIIMS-2004
Dual nature of radiation and Matter
142277
Two identical metal plates show photoelectric effect by a light of wavelength $\lambda_{1}$ on plate 1 and $\lambda_{2}$ on plate 2 (where $\lambda_{1}=2 \lambda_{2}$ ). The maximum kinetic energy will be-
1 $2 \mathrm{~K}_{2}=\mathrm{K}_{1}$
2 $\mathrm{K}_{1} \lt \mathrm{K}_{2} / 2$
3 $\mathrm{K}_{1}>\mathrm{K}_{2} / 2$
4 $2 \mathrm{~K}_{1}=\mathrm{K}_{2}$
Explanation:
B Maximum kinetic energy, $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Condition I- $\because \quad \lambda_{1} =2 \lambda_{2}$ $\mathrm{~K}_{1} =\frac{\mathrm{hc}}{2 \lambda_{2}}-\phi_{\mathrm{o}}$ Condition II- $\because \quad \lambda=\lambda_{2}$ $\mathrm{~K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Dividing both side of the above equation by 2 , $\frac{\mathrm{K}_{2}}{2}=\frac{\mathrm{hc}}{2 \lambda_{2}}-\frac{\phi_{\mathrm{o}}}{2}$ From equation (i) and (ii), we get- $\frac{\mathrm{K}_{2}}{2}-\mathrm{K}_{1}=\frac{\phi_{\mathrm{o}}}{2}$ $\frac{\mathrm{K}_{2}}{2}=\mathrm{K}_{1}+\frac{\phi_{\mathrm{o}}}{2}$ So, from equation (iii) we can say, $\frac{\mathrm{K}_{2}}{2}>\mathrm{K}_{1}$
BCECE-2016
Dual nature of radiation and Matter
142278
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 \mathrm{~cm}^{2}$, then the total momentum delivered on the surface during a span of 30 min is-
A Given, $\mathrm{I}=18 \mathrm{~W} / \mathrm{cm}^{2}, \mathrm{~A}=20 \mathrm{~cm}^{2}, \Delta \mathrm{t}=30$ $\min =30 \times 60=1800 \mathrm{sec}$ Total energy falling on the surface $\mathrm{U}=\mathrm{IA} \Delta \mathrm{t}$ $\mathrm{U}=18 \times 20 \times 1800 \mathrm{~J}$ $\mathrm{U}=6.48 \times 10^{5} \mathrm{~J}$ Momentum, $\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}$ $\mathrm{p}=\frac{6.48 \times 10^{5}}{3 \times 10^{8}}$ $\mathrm{p}=2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
BCECE-2014
Dual nature of radiation and Matter
142279
No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 \mathrm{~nm}$. The work function of the metal is approximately equal to-
1 $3 \times 10^{-16} \mathrm{~J}$
2 $3 \times 10^{-19} \mathrm{~J}$
3 $3 \times 10^{-20} \mathrm{~J}$
4 $3 \times 10^{-22} \mathrm{~J}$
Explanation:
B Given that, $\lambda_{\mathrm{o}}=6000 \stackrel{\circ}{\mathrm{A}}=6 \times 10^{-7} \mathrm{~m}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ Since, we know that work function is given as- $\phi_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}}$ $\phi_{\mathrm{o}}=3 \times 10^{-19} \mathrm{~J}$
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Dual nature of radiation and Matter
142274
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$. Then the work function of the metal must be
1 $\frac{h v}{2}$
2 hv
3 $2 \mathrm{hv}$
4 none of the above
Explanation:
A We know that, $\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ Now, the frequency is increased to $\frac{3 \mathrm{v}}{2}$ $\frac{3 \mathrm{hv}}{2}=\phi+2 \mathrm{~V}_{\mathrm{s}}$ From equation (i) and (ii) we get - $\phi=\frac{\mathrm{hv}}{2}$ Work function $=\frac{\mathrm{hv}}{2}$
AIIMS-2009
Dual nature of radiation and Matter
142275
A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
A Given, $\phi=2 \mathrm{eV}, \mathrm{h} v=4 \mathrm{eV}$ According to Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{h} v-\phi$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{eV}-2 \mathrm{eV}$ Stopping potential- $\mathrm{eV}_{0}=2 \mathrm{eV}$ $\mathrm{V}_{0}=2 \mathrm{~V}$
AIIMS-2004
Dual nature of radiation and Matter
142277
Two identical metal plates show photoelectric effect by a light of wavelength $\lambda_{1}$ on plate 1 and $\lambda_{2}$ on plate 2 (where $\lambda_{1}=2 \lambda_{2}$ ). The maximum kinetic energy will be-
1 $2 \mathrm{~K}_{2}=\mathrm{K}_{1}$
2 $\mathrm{K}_{1} \lt \mathrm{K}_{2} / 2$
3 $\mathrm{K}_{1}>\mathrm{K}_{2} / 2$
4 $2 \mathrm{~K}_{1}=\mathrm{K}_{2}$
Explanation:
B Maximum kinetic energy, $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Condition I- $\because \quad \lambda_{1} =2 \lambda_{2}$ $\mathrm{~K}_{1} =\frac{\mathrm{hc}}{2 \lambda_{2}}-\phi_{\mathrm{o}}$ Condition II- $\because \quad \lambda=\lambda_{2}$ $\mathrm{~K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Dividing both side of the above equation by 2 , $\frac{\mathrm{K}_{2}}{2}=\frac{\mathrm{hc}}{2 \lambda_{2}}-\frac{\phi_{\mathrm{o}}}{2}$ From equation (i) and (ii), we get- $\frac{\mathrm{K}_{2}}{2}-\mathrm{K}_{1}=\frac{\phi_{\mathrm{o}}}{2}$ $\frac{\mathrm{K}_{2}}{2}=\mathrm{K}_{1}+\frac{\phi_{\mathrm{o}}}{2}$ So, from equation (iii) we can say, $\frac{\mathrm{K}_{2}}{2}>\mathrm{K}_{1}$
BCECE-2016
Dual nature of radiation and Matter
142278
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 \mathrm{~cm}^{2}$, then the total momentum delivered on the surface during a span of 30 min is-
A Given, $\mathrm{I}=18 \mathrm{~W} / \mathrm{cm}^{2}, \mathrm{~A}=20 \mathrm{~cm}^{2}, \Delta \mathrm{t}=30$ $\min =30 \times 60=1800 \mathrm{sec}$ Total energy falling on the surface $\mathrm{U}=\mathrm{IA} \Delta \mathrm{t}$ $\mathrm{U}=18 \times 20 \times 1800 \mathrm{~J}$ $\mathrm{U}=6.48 \times 10^{5} \mathrm{~J}$ Momentum, $\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}$ $\mathrm{p}=\frac{6.48 \times 10^{5}}{3 \times 10^{8}}$ $\mathrm{p}=2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
BCECE-2014
Dual nature of radiation and Matter
142279
No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 \mathrm{~nm}$. The work function of the metal is approximately equal to-
1 $3 \times 10^{-16} \mathrm{~J}$
2 $3 \times 10^{-19} \mathrm{~J}$
3 $3 \times 10^{-20} \mathrm{~J}$
4 $3 \times 10^{-22} \mathrm{~J}$
Explanation:
B Given that, $\lambda_{\mathrm{o}}=6000 \stackrel{\circ}{\mathrm{A}}=6 \times 10^{-7} \mathrm{~m}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ Since, we know that work function is given as- $\phi_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}}$ $\phi_{\mathrm{o}}=3 \times 10^{-19} \mathrm{~J}$
142274
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3 v}{2}$. Then the work function of the metal must be
1 $\frac{h v}{2}$
2 hv
3 $2 \mathrm{hv}$
4 none of the above
Explanation:
A We know that, $\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ Now, the frequency is increased to $\frac{3 \mathrm{v}}{2}$ $\frac{3 \mathrm{hv}}{2}=\phi+2 \mathrm{~V}_{\mathrm{s}}$ From equation (i) and (ii) we get - $\phi=\frac{\mathrm{hv}}{2}$ Work function $=\frac{\mathrm{hv}}{2}$
AIIMS-2009
Dual nature of radiation and Matter
142275
A photon of energy $4 \mathrm{eV}$ is incident on a metal surface whose work function is $2 \mathrm{eV}$. The minimum reverse potential to be applied for stopping the emission of electrons is
1 $2 \mathrm{~V}$
2 $4 \mathrm{~V}$
3 $6 \mathrm{~V}$
4 $8 \mathrm{~V}$
Explanation:
A Given, $\phi=2 \mathrm{eV}, \mathrm{h} v=4 \mathrm{eV}$ According to Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=\mathrm{h} v-\phi$ $\frac{1}{2} \mathrm{mv}_{\text {max }}^{2}=4 \mathrm{eV}-2 \mathrm{eV}$ Stopping potential- $\mathrm{eV}_{0}=2 \mathrm{eV}$ $\mathrm{V}_{0}=2 \mathrm{~V}$
AIIMS-2004
Dual nature of radiation and Matter
142277
Two identical metal plates show photoelectric effect by a light of wavelength $\lambda_{1}$ on plate 1 and $\lambda_{2}$ on plate 2 (where $\lambda_{1}=2 \lambda_{2}$ ). The maximum kinetic energy will be-
1 $2 \mathrm{~K}_{2}=\mathrm{K}_{1}$
2 $\mathrm{K}_{1} \lt \mathrm{K}_{2} / 2$
3 $\mathrm{K}_{1}>\mathrm{K}_{2} / 2$
4 $2 \mathrm{~K}_{1}=\mathrm{K}_{2}$
Explanation:
B Maximum kinetic energy, $\mathrm{K}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Condition I- $\because \quad \lambda_{1} =2 \lambda_{2}$ $\mathrm{~K}_{1} =\frac{\mathrm{hc}}{2 \lambda_{2}}-\phi_{\mathrm{o}}$ Condition II- $\because \quad \lambda=\lambda_{2}$ $\mathrm{~K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Dividing both side of the above equation by 2 , $\frac{\mathrm{K}_{2}}{2}=\frac{\mathrm{hc}}{2 \lambda_{2}}-\frac{\phi_{\mathrm{o}}}{2}$ From equation (i) and (ii), we get- $\frac{\mathrm{K}_{2}}{2}-\mathrm{K}_{1}=\frac{\phi_{\mathrm{o}}}{2}$ $\frac{\mathrm{K}_{2}}{2}=\mathrm{K}_{1}+\frac{\phi_{\mathrm{o}}}{2}$ So, from equation (iii) we can say, $\frac{\mathrm{K}_{2}}{2}>\mathrm{K}_{1}$
BCECE-2016
Dual nature of radiation and Matter
142278
Light with an energy flux of $18 \mathrm{~W} / \mathrm{cm}^{2}$ falls on a non-reflecting surface at normal incidence. The surface has an area of $20 \mathrm{~cm}^{2}$, then the total momentum delivered on the surface during a span of 30 min is-
A Given, $\mathrm{I}=18 \mathrm{~W} / \mathrm{cm}^{2}, \mathrm{~A}=20 \mathrm{~cm}^{2}, \Delta \mathrm{t}=30$ $\min =30 \times 60=1800 \mathrm{sec}$ Total energy falling on the surface $\mathrm{U}=\mathrm{IA} \Delta \mathrm{t}$ $\mathrm{U}=18 \times 20 \times 1800 \mathrm{~J}$ $\mathrm{U}=6.48 \times 10^{5} \mathrm{~J}$ Momentum, $\mathrm{p}=\frac{\mathrm{U}}{\mathrm{c}}$ $\mathrm{p}=\frac{6.48 \times 10^{5}}{3 \times 10^{8}}$ $\mathrm{p}=2.16 \times 10^{-3} \mathrm{kgms}^{-1}$
BCECE-2014
Dual nature of radiation and Matter
142279
No photoelectrons are emitted from a metal if the wavelength of the light exceeds $600 \mathrm{~nm}$. The work function of the metal is approximately equal to-
1 $3 \times 10^{-16} \mathrm{~J}$
2 $3 \times 10^{-19} \mathrm{~J}$
3 $3 \times 10^{-20} \mathrm{~J}$
4 $3 \times 10^{-22} \mathrm{~J}$
Explanation:
B Given that, $\lambda_{\mathrm{o}}=6000 \stackrel{\circ}{\mathrm{A}}=6 \times 10^{-7} \mathrm{~m}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}$ Since, we know that work function is given as- $\phi_{\mathrm{o}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\phi_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7}}$ $\phi_{\mathrm{o}}=3 \times 10^{-19} \mathrm{~J}$