142115
The threshold wavelength for a metal is 5200 $\AA$. The photoelectrons will be ejected if it is irradiated by light from a
1 54 watt infrared lamp
2 1 watt infrared lamp
3 50 watt ultraviolet lamp
4 0.5 watt infrared lamp
Explanation:
C Given that, Threshold wavelength $=5200 \AA$ For photoelectric emission the wavelength of incident light should be smaller than threshold wavelength. The wavelength of ultraviolet light is about $4000 \AA$. And infrared region has wavelength greater than 8000 Å. Hence, infrared lamp is not useful for the purpose. The 50 watt ultra violet lamp can be used.
A According to the Einstein's photoelectric equation is - $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{h}=$ Planck's constant $v=\text { frequency of incident radiation }$ $\phi=\text { work function }$ $(\mathrm{KE})_{\max }=\text { maximum kinetic energy }$
J and K CET-2013
Dual nature of radiation and Matter
142134
The threshold frequency of a photoelectric metal is $v_{0}$. If light of frequency $4 v_{0}$ is incident on this metal then the maximum kinetic energy of emitted electrons will be:
1 $h v_{\mathrm{o}}$
2 $2 \mathrm{~h} v_{\mathrm{o}}$
3 $3 \mathrm{~h} v_{\mathrm{o}}$
4 $4 \mathrm{~h} v_{\mathrm{o}}$
Explanation:
C We know that $(\mathrm{KE})_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ $=\mathrm{h}\left(4 v_{\mathrm{o}}\right)-\mathrm{h} v_{\mathrm{o}}$ $=3 \mathrm{~h} v_{\mathrm{o}}$
Dual nature of radiation and Matter
142137
Maximum kinetic energy of electrons emitted in photoelectric effect increases when :
1 intensity of light is increased
2 light source is brought nearer the metal
3 frequency of light is decreased
4 wavelength of light is decreased
Explanation:
D According to the Einstein photoelectric equation- $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Where, $\phi_{\mathrm{o}}$ is constant Hence, KE is inversely proportional to $\lambda$ as wavelength is decreased the maximum kinetic energy is increased.
142115
The threshold wavelength for a metal is 5200 $\AA$. The photoelectrons will be ejected if it is irradiated by light from a
1 54 watt infrared lamp
2 1 watt infrared lamp
3 50 watt ultraviolet lamp
4 0.5 watt infrared lamp
Explanation:
C Given that, Threshold wavelength $=5200 \AA$ For photoelectric emission the wavelength of incident light should be smaller than threshold wavelength. The wavelength of ultraviolet light is about $4000 \AA$. And infrared region has wavelength greater than 8000 Å. Hence, infrared lamp is not useful for the purpose. The 50 watt ultra violet lamp can be used.
A According to the Einstein's photoelectric equation is - $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{h}=$ Planck's constant $v=\text { frequency of incident radiation }$ $\phi=\text { work function }$ $(\mathrm{KE})_{\max }=\text { maximum kinetic energy }$
J and K CET-2013
Dual nature of radiation and Matter
142134
The threshold frequency of a photoelectric metal is $v_{0}$. If light of frequency $4 v_{0}$ is incident on this metal then the maximum kinetic energy of emitted electrons will be:
1 $h v_{\mathrm{o}}$
2 $2 \mathrm{~h} v_{\mathrm{o}}$
3 $3 \mathrm{~h} v_{\mathrm{o}}$
4 $4 \mathrm{~h} v_{\mathrm{o}}$
Explanation:
C We know that $(\mathrm{KE})_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ $=\mathrm{h}\left(4 v_{\mathrm{o}}\right)-\mathrm{h} v_{\mathrm{o}}$ $=3 \mathrm{~h} v_{\mathrm{o}}$
Dual nature of radiation and Matter
142137
Maximum kinetic energy of electrons emitted in photoelectric effect increases when :
1 intensity of light is increased
2 light source is brought nearer the metal
3 frequency of light is decreased
4 wavelength of light is decreased
Explanation:
D According to the Einstein photoelectric equation- $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Where, $\phi_{\mathrm{o}}$ is constant Hence, KE is inversely proportional to $\lambda$ as wavelength is decreased the maximum kinetic energy is increased.
142115
The threshold wavelength for a metal is 5200 $\AA$. The photoelectrons will be ejected if it is irradiated by light from a
1 54 watt infrared lamp
2 1 watt infrared lamp
3 50 watt ultraviolet lamp
4 0.5 watt infrared lamp
Explanation:
C Given that, Threshold wavelength $=5200 \AA$ For photoelectric emission the wavelength of incident light should be smaller than threshold wavelength. The wavelength of ultraviolet light is about $4000 \AA$. And infrared region has wavelength greater than 8000 Å. Hence, infrared lamp is not useful for the purpose. The 50 watt ultra violet lamp can be used.
A According to the Einstein's photoelectric equation is - $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{h}=$ Planck's constant $v=\text { frequency of incident radiation }$ $\phi=\text { work function }$ $(\mathrm{KE})_{\max }=\text { maximum kinetic energy }$
J and K CET-2013
Dual nature of radiation and Matter
142134
The threshold frequency of a photoelectric metal is $v_{0}$. If light of frequency $4 v_{0}$ is incident on this metal then the maximum kinetic energy of emitted electrons will be:
1 $h v_{\mathrm{o}}$
2 $2 \mathrm{~h} v_{\mathrm{o}}$
3 $3 \mathrm{~h} v_{\mathrm{o}}$
4 $4 \mathrm{~h} v_{\mathrm{o}}$
Explanation:
C We know that $(\mathrm{KE})_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ $=\mathrm{h}\left(4 v_{\mathrm{o}}\right)-\mathrm{h} v_{\mathrm{o}}$ $=3 \mathrm{~h} v_{\mathrm{o}}$
Dual nature of radiation and Matter
142137
Maximum kinetic energy of electrons emitted in photoelectric effect increases when :
1 intensity of light is increased
2 light source is brought nearer the metal
3 frequency of light is decreased
4 wavelength of light is decreased
Explanation:
D According to the Einstein photoelectric equation- $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Where, $\phi_{\mathrm{o}}$ is constant Hence, KE is inversely proportional to $\lambda$ as wavelength is decreased the maximum kinetic energy is increased.
142115
The threshold wavelength for a metal is 5200 $\AA$. The photoelectrons will be ejected if it is irradiated by light from a
1 54 watt infrared lamp
2 1 watt infrared lamp
3 50 watt ultraviolet lamp
4 0.5 watt infrared lamp
Explanation:
C Given that, Threshold wavelength $=5200 \AA$ For photoelectric emission the wavelength of incident light should be smaller than threshold wavelength. The wavelength of ultraviolet light is about $4000 \AA$. And infrared region has wavelength greater than 8000 Å. Hence, infrared lamp is not useful for the purpose. The 50 watt ultra violet lamp can be used.
A According to the Einstein's photoelectric equation is - $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi$ Where, $\mathrm{h}=$ Planck's constant $v=\text { frequency of incident radiation }$ $\phi=\text { work function }$ $(\mathrm{KE})_{\max }=\text { maximum kinetic energy }$
J and K CET-2013
Dual nature of radiation and Matter
142134
The threshold frequency of a photoelectric metal is $v_{0}$. If light of frequency $4 v_{0}$ is incident on this metal then the maximum kinetic energy of emitted electrons will be:
1 $h v_{\mathrm{o}}$
2 $2 \mathrm{~h} v_{\mathrm{o}}$
3 $3 \mathrm{~h} v_{\mathrm{o}}$
4 $4 \mathrm{~h} v_{\mathrm{o}}$
Explanation:
C We know that $(\mathrm{KE})_{\max }=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ $=\mathrm{h}\left(4 v_{\mathrm{o}}\right)-\mathrm{h} v_{\mathrm{o}}$ $=3 \mathrm{~h} v_{\mathrm{o}}$
Dual nature of radiation and Matter
142137
Maximum kinetic energy of electrons emitted in photoelectric effect increases when :
1 intensity of light is increased
2 light source is brought nearer the metal
3 frequency of light is decreased
4 wavelength of light is decreased
Explanation:
D According to the Einstein photoelectric equation- $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Where, $\phi_{\mathrm{o}}$ is constant Hence, KE is inversely proportional to $\lambda$ as wavelength is decreased the maximum kinetic energy is increased.
