142223
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum K.E. of the emitted photoelectrons would be
1 four times the original value
2 unchanged
3 twice the original value
4 one fourth of the original value
Explanation:
B According to Einstein's equation for photoelectric effect $\mathrm{K}_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ $(\mathrm{K} . \mathrm{E})_{\text {max }}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}}$ $\mathrm{h}=$ Planck's constant, $v_{\mathrm{o}}=$ threshold frequency $(\mathrm{K} . \mathrm{E}) \propto v$ The (K.E $)_{\max }$ of emitted photo electron is proportional to the frequency of radiation and is independent of the intensity of radiation so it's remains unchanged.
MHT-CET 2019
Dual nature of radiation and Matter
142235
A monochromatic light of frequency $v$ is incident on emitter having threshold frequency $v_{0}$. The kinetic energy of ejected electron will be
B According to Einstein equation for photoelectric $\mathrm{E}=\phi+\mathrm{KE} \quad\left(\because \phi=h v_{\mathrm{o}}\right)$ $\mathrm{KE}_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}(\because \mathrm{E}=\mathrm{h} v)$ $\mathrm{KE}_{\max }=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\mathrm{h}=$ Planck's constant $v_{\mathrm{o}}=\text { Threshold frequency }$
2011
Dual nature of radiation and Matter
142244
In Einstein's photoelectric equation $\mathrm{E}_{\mathrm{kin}}=\mathrm{hv}-\phi$, $E_{\text {kin }}$ stands for
1 kinetic energy of all the emitted electrons
2 mean kinetic energy of the emitted electrons
3 maximum kinetic energy of the emitted electron
4 minimum kinetic energy of the emitted electrons
Explanation:
C Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{kin}}=\mathrm{h} v-\phi$ Where, $\phi=$ Work function $h v=$ Energy of photon $\mathrm{E}_{\mathrm{k}}=$ Maximum kinetic energy of the emitted electron
CG PET- 2004
Dual nature of radiation and Matter
142247
The number of photoelectrons in photoelectric effect experiment depends on
1 Frequency of light
2 Intensity of light
3 Both (a) and (b) are correct
4 Both (a) and (b) are incorrect
Explanation:
B The number of photoelectrons in photoelectric effect depends on intensity of light only. $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{E}_{\mathrm{P}}}$ $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{h} v}$ $\mathrm{E}_{\mathrm{P}}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ $\left(\because v>v_{0}\right)$
BITSAT-2013
Dual nature of radiation and Matter
142251
The energy of a photon of wavelength $\lambda$ is
1 hc $\lambda$
2 $\frac{\mathrm{hc}}{\lambda}$
3 $\frac{\lambda}{\mathrm{hc}}$
4 $\frac{\mathrm{h} \lambda}{\mathrm{c}}$
Explanation:
B Photon Energy $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda} \quad\left(\because v=\frac{\mathrm{c}}{\lambda}\right)$ Where, $\mathrm{E}=$ Photon energy $\mathrm{h}=\text { Planck constant }$ $\mathrm{c}=\text { Speed of light }$ $\lambda=\text { Photon wavelength }$ $\mathrm{v}=\text { Photon frequency }$
142223
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum K.E. of the emitted photoelectrons would be
1 four times the original value
2 unchanged
3 twice the original value
4 one fourth of the original value
Explanation:
B According to Einstein's equation for photoelectric effect $\mathrm{K}_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ $(\mathrm{K} . \mathrm{E})_{\text {max }}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}}$ $\mathrm{h}=$ Planck's constant, $v_{\mathrm{o}}=$ threshold frequency $(\mathrm{K} . \mathrm{E}) \propto v$ The (K.E $)_{\max }$ of emitted photo electron is proportional to the frequency of radiation and is independent of the intensity of radiation so it's remains unchanged.
MHT-CET 2019
Dual nature of radiation and Matter
142235
A monochromatic light of frequency $v$ is incident on emitter having threshold frequency $v_{0}$. The kinetic energy of ejected electron will be
B According to Einstein equation for photoelectric $\mathrm{E}=\phi+\mathrm{KE} \quad\left(\because \phi=h v_{\mathrm{o}}\right)$ $\mathrm{KE}_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}(\because \mathrm{E}=\mathrm{h} v)$ $\mathrm{KE}_{\max }=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\mathrm{h}=$ Planck's constant $v_{\mathrm{o}}=\text { Threshold frequency }$
2011
Dual nature of radiation and Matter
142244
In Einstein's photoelectric equation $\mathrm{E}_{\mathrm{kin}}=\mathrm{hv}-\phi$, $E_{\text {kin }}$ stands for
1 kinetic energy of all the emitted electrons
2 mean kinetic energy of the emitted electrons
3 maximum kinetic energy of the emitted electron
4 minimum kinetic energy of the emitted electrons
Explanation:
C Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{kin}}=\mathrm{h} v-\phi$ Where, $\phi=$ Work function $h v=$ Energy of photon $\mathrm{E}_{\mathrm{k}}=$ Maximum kinetic energy of the emitted electron
CG PET- 2004
Dual nature of radiation and Matter
142247
The number of photoelectrons in photoelectric effect experiment depends on
1 Frequency of light
2 Intensity of light
3 Both (a) and (b) are correct
4 Both (a) and (b) are incorrect
Explanation:
B The number of photoelectrons in photoelectric effect depends on intensity of light only. $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{E}_{\mathrm{P}}}$ $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{h} v}$ $\mathrm{E}_{\mathrm{P}}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ $\left(\because v>v_{0}\right)$
BITSAT-2013
Dual nature of radiation and Matter
142251
The energy of a photon of wavelength $\lambda$ is
1 hc $\lambda$
2 $\frac{\mathrm{hc}}{\lambda}$
3 $\frac{\lambda}{\mathrm{hc}}$
4 $\frac{\mathrm{h} \lambda}{\mathrm{c}}$
Explanation:
B Photon Energy $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda} \quad\left(\because v=\frac{\mathrm{c}}{\lambda}\right)$ Where, $\mathrm{E}=$ Photon energy $\mathrm{h}=\text { Planck constant }$ $\mathrm{c}=\text { Speed of light }$ $\lambda=\text { Photon wavelength }$ $\mathrm{v}=\text { Photon frequency }$
142223
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum K.E. of the emitted photoelectrons would be
1 four times the original value
2 unchanged
3 twice the original value
4 one fourth of the original value
Explanation:
B According to Einstein's equation for photoelectric effect $\mathrm{K}_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ $(\mathrm{K} . \mathrm{E})_{\text {max }}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}}$ $\mathrm{h}=$ Planck's constant, $v_{\mathrm{o}}=$ threshold frequency $(\mathrm{K} . \mathrm{E}) \propto v$ The (K.E $)_{\max }$ of emitted photo electron is proportional to the frequency of radiation and is independent of the intensity of radiation so it's remains unchanged.
MHT-CET 2019
Dual nature of radiation and Matter
142235
A monochromatic light of frequency $v$ is incident on emitter having threshold frequency $v_{0}$. The kinetic energy of ejected electron will be
B According to Einstein equation for photoelectric $\mathrm{E}=\phi+\mathrm{KE} \quad\left(\because \phi=h v_{\mathrm{o}}\right)$ $\mathrm{KE}_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}(\because \mathrm{E}=\mathrm{h} v)$ $\mathrm{KE}_{\max }=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\mathrm{h}=$ Planck's constant $v_{\mathrm{o}}=\text { Threshold frequency }$
2011
Dual nature of radiation and Matter
142244
In Einstein's photoelectric equation $\mathrm{E}_{\mathrm{kin}}=\mathrm{hv}-\phi$, $E_{\text {kin }}$ stands for
1 kinetic energy of all the emitted electrons
2 mean kinetic energy of the emitted electrons
3 maximum kinetic energy of the emitted electron
4 minimum kinetic energy of the emitted electrons
Explanation:
C Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{kin}}=\mathrm{h} v-\phi$ Where, $\phi=$ Work function $h v=$ Energy of photon $\mathrm{E}_{\mathrm{k}}=$ Maximum kinetic energy of the emitted electron
CG PET- 2004
Dual nature of radiation and Matter
142247
The number of photoelectrons in photoelectric effect experiment depends on
1 Frequency of light
2 Intensity of light
3 Both (a) and (b) are correct
4 Both (a) and (b) are incorrect
Explanation:
B The number of photoelectrons in photoelectric effect depends on intensity of light only. $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{E}_{\mathrm{P}}}$ $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{h} v}$ $\mathrm{E}_{\mathrm{P}}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ $\left(\because v>v_{0}\right)$
BITSAT-2013
Dual nature of radiation and Matter
142251
The energy of a photon of wavelength $\lambda$ is
1 hc $\lambda$
2 $\frac{\mathrm{hc}}{\lambda}$
3 $\frac{\lambda}{\mathrm{hc}}$
4 $\frac{\mathrm{h} \lambda}{\mathrm{c}}$
Explanation:
B Photon Energy $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda} \quad\left(\because v=\frac{\mathrm{c}}{\lambda}\right)$ Where, $\mathrm{E}=$ Photon energy $\mathrm{h}=\text { Planck constant }$ $\mathrm{c}=\text { Speed of light }$ $\lambda=\text { Photon wavelength }$ $\mathrm{v}=\text { Photon frequency }$
142223
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum K.E. of the emitted photoelectrons would be
1 four times the original value
2 unchanged
3 twice the original value
4 one fourth of the original value
Explanation:
B According to Einstein's equation for photoelectric effect $\mathrm{K}_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ $(\mathrm{K} . \mathrm{E})_{\text {max }}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}}$ $\mathrm{h}=$ Planck's constant, $v_{\mathrm{o}}=$ threshold frequency $(\mathrm{K} . \mathrm{E}) \propto v$ The (K.E $)_{\max }$ of emitted photo electron is proportional to the frequency of radiation and is independent of the intensity of radiation so it's remains unchanged.
MHT-CET 2019
Dual nature of radiation and Matter
142235
A monochromatic light of frequency $v$ is incident on emitter having threshold frequency $v_{0}$. The kinetic energy of ejected electron will be
B According to Einstein equation for photoelectric $\mathrm{E}=\phi+\mathrm{KE} \quad\left(\because \phi=h v_{\mathrm{o}}\right)$ $\mathrm{KE}_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}(\because \mathrm{E}=\mathrm{h} v)$ $\mathrm{KE}_{\max }=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\mathrm{h}=$ Planck's constant $v_{\mathrm{o}}=\text { Threshold frequency }$
2011
Dual nature of radiation and Matter
142244
In Einstein's photoelectric equation $\mathrm{E}_{\mathrm{kin}}=\mathrm{hv}-\phi$, $E_{\text {kin }}$ stands for
1 kinetic energy of all the emitted electrons
2 mean kinetic energy of the emitted electrons
3 maximum kinetic energy of the emitted electron
4 minimum kinetic energy of the emitted electrons
Explanation:
C Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{kin}}=\mathrm{h} v-\phi$ Where, $\phi=$ Work function $h v=$ Energy of photon $\mathrm{E}_{\mathrm{k}}=$ Maximum kinetic energy of the emitted electron
CG PET- 2004
Dual nature of radiation and Matter
142247
The number of photoelectrons in photoelectric effect experiment depends on
1 Frequency of light
2 Intensity of light
3 Both (a) and (b) are correct
4 Both (a) and (b) are incorrect
Explanation:
B The number of photoelectrons in photoelectric effect depends on intensity of light only. $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{E}_{\mathrm{P}}}$ $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{h} v}$ $\mathrm{E}_{\mathrm{P}}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ $\left(\because v>v_{0}\right)$
BITSAT-2013
Dual nature of radiation and Matter
142251
The energy of a photon of wavelength $\lambda$ is
1 hc $\lambda$
2 $\frac{\mathrm{hc}}{\lambda}$
3 $\frac{\lambda}{\mathrm{hc}}$
4 $\frac{\mathrm{h} \lambda}{\mathrm{c}}$
Explanation:
B Photon Energy $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda} \quad\left(\because v=\frac{\mathrm{c}}{\lambda}\right)$ Where, $\mathrm{E}=$ Photon energy $\mathrm{h}=\text { Planck constant }$ $\mathrm{c}=\text { Speed of light }$ $\lambda=\text { Photon wavelength }$ $\mathrm{v}=\text { Photon frequency }$
142223
A metal surface is illuminated by light of given intensity and frequency to cause photoemission. If the intensity of illumination is reduced to one fourth of its original value then the maximum K.E. of the emitted photoelectrons would be
1 four times the original value
2 unchanged
3 twice the original value
4 one fourth of the original value
Explanation:
B According to Einstein's equation for photoelectric effect $\mathrm{K}_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ $(\mathrm{K} . \mathrm{E})_{\text {max }}=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\phi_{\mathrm{o}}=\mathrm{h} \mathrm{v}_{\mathrm{o}}$ $\mathrm{h}=$ Planck's constant, $v_{\mathrm{o}}=$ threshold frequency $(\mathrm{K} . \mathrm{E}) \propto v$ The (K.E $)_{\max }$ of emitted photo electron is proportional to the frequency of radiation and is independent of the intensity of radiation so it's remains unchanged.
MHT-CET 2019
Dual nature of radiation and Matter
142235
A monochromatic light of frequency $v$ is incident on emitter having threshold frequency $v_{0}$. The kinetic energy of ejected electron will be
B According to Einstein equation for photoelectric $\mathrm{E}=\phi+\mathrm{KE} \quad\left(\because \phi=h v_{\mathrm{o}}\right)$ $\mathrm{KE}_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}(\because \mathrm{E}=\mathrm{h} v)$ $\mathrm{KE}_{\max }=\mathrm{h}\left(v-v_{\mathrm{o}}\right)$ Where, $\mathrm{h}=$ Planck's constant $v_{\mathrm{o}}=\text { Threshold frequency }$
2011
Dual nature of radiation and Matter
142244
In Einstein's photoelectric equation $\mathrm{E}_{\mathrm{kin}}=\mathrm{hv}-\phi$, $E_{\text {kin }}$ stands for
1 kinetic energy of all the emitted electrons
2 mean kinetic energy of the emitted electrons
3 maximum kinetic energy of the emitted electron
4 minimum kinetic energy of the emitted electrons
Explanation:
C Einstein's photoelectric equation- $\mathrm{E}_{\mathrm{kin}}=\mathrm{h} v-\phi$ Where, $\phi=$ Work function $h v=$ Energy of photon $\mathrm{E}_{\mathrm{k}}=$ Maximum kinetic energy of the emitted electron
CG PET- 2004
Dual nature of radiation and Matter
142247
The number of photoelectrons in photoelectric effect experiment depends on
1 Frequency of light
2 Intensity of light
3 Both (a) and (b) are correct
4 Both (a) and (b) are incorrect
Explanation:
B The number of photoelectrons in photoelectric effect depends on intensity of light only. $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{E}_{\mathrm{P}}}$ $\mathrm{n}=\frac{\mathrm{I} \times \mathrm{A}}{\mathrm{h} v}$ $\mathrm{E}_{\mathrm{P}}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ $\left(\because v>v_{0}\right)$
BITSAT-2013
Dual nature of radiation and Matter
142251
The energy of a photon of wavelength $\lambda$ is
1 hc $\lambda$
2 $\frac{\mathrm{hc}}{\lambda}$
3 $\frac{\lambda}{\mathrm{hc}}$
4 $\frac{\mathrm{h} \lambda}{\mathrm{c}}$
Explanation:
B Photon Energy $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda} \quad\left(\because v=\frac{\mathrm{c}}{\lambda}\right)$ Where, $\mathrm{E}=$ Photon energy $\mathrm{h}=\text { Planck constant }$ $\mathrm{c}=\text { Speed of light }$ $\lambda=\text { Photon wavelength }$ $\mathrm{v}=\text { Photon frequency }$
