142295
If the work function of a metal is $\phi$ and $f$ is the frequency incident on it then there is no photoelectric emission when
1 $\phi=\frac{\mathrm{hf}}{\mathrm{c}}$
2 $\phi>\frac{\mathrm{hf}}{\mathrm{c}}$
3 $\phi \lt \mathrm{hf}$
4 $\phi>\mathrm{hf}$
Explanation:
D For no emission of photo-electron energy of incident $ \lt $ work function $\text { hf } \lt \phi$ Where, $\mathrm{h}=$ Planck's constant $\mathrm{f}=$ frequency $\phi=$ work function
J and K CET- 1998
Dual nature of radiation and Matter
142298
Which one of the following equation is Einstein's equation related to photoelectric effect?
B Einstein's photo-electric equation, $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{h} v-\mathrm{W}$ Energy of incident radiation $=h v$ $\frac{1}{2} \mathrm{mv}^{2}+\mathrm{W}=\mathrm{hv}$ Where, $\mathrm{W}$ is the work function.
WB JEE-2007
Dual nature of radiation and Matter
142302
If we consider electrons and photons of the same wavelength, they will have the same
1 velocity
2 energy
3 momentum
4 angular momentum
Explanation:
C We know that, $\lambda_{\text {photons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of proton, $\mathrm{p}=\frac{\mathrm{h}}{\lambda_{\text {photons }}}$ $\lambda_{\text {electrons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of electron, $\qquad \mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\text {electrons }}}$ $\text { And } \lambda_{\text {photons }}=\lambda_{\text {electrons }}$ $\text { Therefore }\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {photons }}=\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {electrons }}$ $\therefore \quad \mathrm{p}_{\text {photons }}=\mathrm{p}_{\text {electrons }}$ So, they will have the same momentum.
SRMJEEE - 2014
Dual nature of radiation and Matter
142326
When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{\text {mas }}$. If the intensity of incident light is doubled, then
1 both $n$ and $K_{\max }$ are halved
2 $\mathrm{n}$ is doubled but $\mathrm{K}_{\max }$ remains same
3 $\mathrm{K}_{\max }$ is doubled but $\mathrm{n}$ remains same
4 both $\mathrm{n}$ and $\mathrm{K}_{\max }$ are doubled
Explanation:
B The maximum kinetic energy of emitted photoelectrons $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{K}_{\max }$ represents the maximum kinetic energy $\mathrm{h}=\text { Planck's constant }$ $v=$ frequency It is dependent mainly on the energy of the incident photon on doubling the intensity $\mathrm{K}_{\max }$ remains the same. Intensity is proportional to number of photon per second. Hence, if the intensity is doubled then number of photoelectron emitted per second is also doubled.
142295
If the work function of a metal is $\phi$ and $f$ is the frequency incident on it then there is no photoelectric emission when
1 $\phi=\frac{\mathrm{hf}}{\mathrm{c}}$
2 $\phi>\frac{\mathrm{hf}}{\mathrm{c}}$
3 $\phi \lt \mathrm{hf}$
4 $\phi>\mathrm{hf}$
Explanation:
D For no emission of photo-electron energy of incident $ \lt $ work function $\text { hf } \lt \phi$ Where, $\mathrm{h}=$ Planck's constant $\mathrm{f}=$ frequency $\phi=$ work function
J and K CET- 1998
Dual nature of radiation and Matter
142298
Which one of the following equation is Einstein's equation related to photoelectric effect?
B Einstein's photo-electric equation, $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{h} v-\mathrm{W}$ Energy of incident radiation $=h v$ $\frac{1}{2} \mathrm{mv}^{2}+\mathrm{W}=\mathrm{hv}$ Where, $\mathrm{W}$ is the work function.
WB JEE-2007
Dual nature of radiation and Matter
142302
If we consider electrons and photons of the same wavelength, they will have the same
1 velocity
2 energy
3 momentum
4 angular momentum
Explanation:
C We know that, $\lambda_{\text {photons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of proton, $\mathrm{p}=\frac{\mathrm{h}}{\lambda_{\text {photons }}}$ $\lambda_{\text {electrons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of electron, $\qquad \mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\text {electrons }}}$ $\text { And } \lambda_{\text {photons }}=\lambda_{\text {electrons }}$ $\text { Therefore }\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {photons }}=\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {electrons }}$ $\therefore \quad \mathrm{p}_{\text {photons }}=\mathrm{p}_{\text {electrons }}$ So, they will have the same momentum.
SRMJEEE - 2014
Dual nature of radiation and Matter
142326
When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{\text {mas }}$. If the intensity of incident light is doubled, then
1 both $n$ and $K_{\max }$ are halved
2 $\mathrm{n}$ is doubled but $\mathrm{K}_{\max }$ remains same
3 $\mathrm{K}_{\max }$ is doubled but $\mathrm{n}$ remains same
4 both $\mathrm{n}$ and $\mathrm{K}_{\max }$ are doubled
Explanation:
B The maximum kinetic energy of emitted photoelectrons $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{K}_{\max }$ represents the maximum kinetic energy $\mathrm{h}=\text { Planck's constant }$ $v=$ frequency It is dependent mainly on the energy of the incident photon on doubling the intensity $\mathrm{K}_{\max }$ remains the same. Intensity is proportional to number of photon per second. Hence, if the intensity is doubled then number of photoelectron emitted per second is also doubled.
142295
If the work function of a metal is $\phi$ and $f$ is the frequency incident on it then there is no photoelectric emission when
1 $\phi=\frac{\mathrm{hf}}{\mathrm{c}}$
2 $\phi>\frac{\mathrm{hf}}{\mathrm{c}}$
3 $\phi \lt \mathrm{hf}$
4 $\phi>\mathrm{hf}$
Explanation:
D For no emission of photo-electron energy of incident $ \lt $ work function $\text { hf } \lt \phi$ Where, $\mathrm{h}=$ Planck's constant $\mathrm{f}=$ frequency $\phi=$ work function
J and K CET- 1998
Dual nature of radiation and Matter
142298
Which one of the following equation is Einstein's equation related to photoelectric effect?
B Einstein's photo-electric equation, $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{h} v-\mathrm{W}$ Energy of incident radiation $=h v$ $\frac{1}{2} \mathrm{mv}^{2}+\mathrm{W}=\mathrm{hv}$ Where, $\mathrm{W}$ is the work function.
WB JEE-2007
Dual nature of radiation and Matter
142302
If we consider electrons and photons of the same wavelength, they will have the same
1 velocity
2 energy
3 momentum
4 angular momentum
Explanation:
C We know that, $\lambda_{\text {photons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of proton, $\mathrm{p}=\frac{\mathrm{h}}{\lambda_{\text {photons }}}$ $\lambda_{\text {electrons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of electron, $\qquad \mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\text {electrons }}}$ $\text { And } \lambda_{\text {photons }}=\lambda_{\text {electrons }}$ $\text { Therefore }\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {photons }}=\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {electrons }}$ $\therefore \quad \mathrm{p}_{\text {photons }}=\mathrm{p}_{\text {electrons }}$ So, they will have the same momentum.
SRMJEEE - 2014
Dual nature of radiation and Matter
142326
When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{\text {mas }}$. If the intensity of incident light is doubled, then
1 both $n$ and $K_{\max }$ are halved
2 $\mathrm{n}$ is doubled but $\mathrm{K}_{\max }$ remains same
3 $\mathrm{K}_{\max }$ is doubled but $\mathrm{n}$ remains same
4 both $\mathrm{n}$ and $\mathrm{K}_{\max }$ are doubled
Explanation:
B The maximum kinetic energy of emitted photoelectrons $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{K}_{\max }$ represents the maximum kinetic energy $\mathrm{h}=\text { Planck's constant }$ $v=$ frequency It is dependent mainly on the energy of the incident photon on doubling the intensity $\mathrm{K}_{\max }$ remains the same. Intensity is proportional to number of photon per second. Hence, if the intensity is doubled then number of photoelectron emitted per second is also doubled.
142295
If the work function of a metal is $\phi$ and $f$ is the frequency incident on it then there is no photoelectric emission when
1 $\phi=\frac{\mathrm{hf}}{\mathrm{c}}$
2 $\phi>\frac{\mathrm{hf}}{\mathrm{c}}$
3 $\phi \lt \mathrm{hf}$
4 $\phi>\mathrm{hf}$
Explanation:
D For no emission of photo-electron energy of incident $ \lt $ work function $\text { hf } \lt \phi$ Where, $\mathrm{h}=$ Planck's constant $\mathrm{f}=$ frequency $\phi=$ work function
J and K CET- 1998
Dual nature of radiation and Matter
142298
Which one of the following equation is Einstein's equation related to photoelectric effect?
B Einstein's photo-electric equation, $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{h} v-\mathrm{W}$ Energy of incident radiation $=h v$ $\frac{1}{2} \mathrm{mv}^{2}+\mathrm{W}=\mathrm{hv}$ Where, $\mathrm{W}$ is the work function.
WB JEE-2007
Dual nature of radiation and Matter
142302
If we consider electrons and photons of the same wavelength, they will have the same
1 velocity
2 energy
3 momentum
4 angular momentum
Explanation:
C We know that, $\lambda_{\text {photons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of proton, $\mathrm{p}=\frac{\mathrm{h}}{\lambda_{\text {photons }}}$ $\lambda_{\text {electrons }}=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum of electron, $\qquad \mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\text {electrons }}}$ $\text { And } \lambda_{\text {photons }}=\lambda_{\text {electrons }}$ $\text { Therefore }\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {photons }}=\left(\frac{\mathrm{h}}{\mathrm{p}}\right)_{\text {electrons }}$ $\therefore \quad \mathrm{p}_{\text {photons }}=\mathrm{p}_{\text {electrons }}$ So, they will have the same momentum.
SRMJEEE - 2014
Dual nature of radiation and Matter
142326
When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is $n$ and their maximum kinetic energy is $K_{\text {mas }}$. If the intensity of incident light is doubled, then
1 both $n$ and $K_{\max }$ are halved
2 $\mathrm{n}$ is doubled but $\mathrm{K}_{\max }$ remains same
3 $\mathrm{K}_{\max }$ is doubled but $\mathrm{n}$ remains same
4 both $\mathrm{n}$ and $\mathrm{K}_{\max }$ are doubled
Explanation:
B The maximum kinetic energy of emitted photoelectrons $\mathrm{K}_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{K}_{\max }$ represents the maximum kinetic energy $\mathrm{h}=\text { Planck's constant }$ $v=$ frequency It is dependent mainly on the energy of the incident photon on doubling the intensity $\mathrm{K}_{\max }$ remains the same. Intensity is proportional to number of photon per second. Hence, if the intensity is doubled then number of photoelectron emitted per second is also doubled.
