142201
Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is
1 $3: 2$
2 $2: 3$
3 $1: 3$
4 $1: 2$
Explanation:
C We know that kinetic energy in photoelectric effect, $\mathrm{K}_{\max }=\mathrm{E}-\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=2 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=10 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=9 \phi_{\mathrm{o}}$ Dividing equation (i) by equation (ii), we get - $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{9}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{3}$ Hence, the maximum velocity ratio - $\mathrm{v}_{1}: \mathrm{v}_{2}=1: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142202
Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm}$. What is the ratio of work functions for lights of wavelength ' $\lambda_{1}$ ' to ' $\lambda_{2}$ '?
1 $6: 1$
2 $3: 5$
3 $5: 3$
4 $1: 6$
Explanation:
C Given that, $\lambda_{1}=360 \mathrm{~nm}, \lambda_{2}=600 \mathrm{~nm}$ Work function - $\phi=\frac{h \mathrm{c}}{\lambda}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{600}{360}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{5}{3}$ $\phi_{1}: \phi_{2}=5: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142203
Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is
1 $1: 4$
2 $1: 2$
3 $1: \sqrt{2}$
4 $1: 1$
Explanation:
C Maximum K.E = Incident photon energy work function $\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W}$ According to question- $\frac{1}{2} \mathrm{mv}_{1}^{2}=3 \mathrm{~W}-\mathrm{W}=2 \mathrm{~W}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=5 \mathrm{~W}-\mathrm{W}=4 \mathrm{~W}$ Dividing equation (i) by equation (ii), we get- $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{2}$
MHT-CET 2020
Dual nature of radiation and Matter
142204
The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $\mathrm{X}$ axis respectively are
1 Planck's constant, threshold frequency
2 work function, maximum K.E.
3 Planck's constant, work function
4 Maximum K.E, threshold frequency
Explanation:
A From Einstein's photoelectric equation, (K.E $)_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ Equation of straight line, Here $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$ Here $\mathrm{m}=\mathrm{h}$ $\mathrm{y}$-intercept $\mathrm{c}=-\phi_{\mathrm{o}}$ So, the slope of graph and intercept on $\mathrm{x}$-axis shown Planck's constant and threshold frequency.
MHT-CET 2020
Dual nature of radiation and Matter
142205
The light of wavelength ' $\lambda$ '. Incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light,$h=$ Planck's constant, $m=$ mass of electron)
142201
Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is
1 $3: 2$
2 $2: 3$
3 $1: 3$
4 $1: 2$
Explanation:
C We know that kinetic energy in photoelectric effect, $\mathrm{K}_{\max }=\mathrm{E}-\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=2 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=10 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=9 \phi_{\mathrm{o}}$ Dividing equation (i) by equation (ii), we get - $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{9}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{3}$ Hence, the maximum velocity ratio - $\mathrm{v}_{1}: \mathrm{v}_{2}=1: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142202
Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm}$. What is the ratio of work functions for lights of wavelength ' $\lambda_{1}$ ' to ' $\lambda_{2}$ '?
1 $6: 1$
2 $3: 5$
3 $5: 3$
4 $1: 6$
Explanation:
C Given that, $\lambda_{1}=360 \mathrm{~nm}, \lambda_{2}=600 \mathrm{~nm}$ Work function - $\phi=\frac{h \mathrm{c}}{\lambda}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{600}{360}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{5}{3}$ $\phi_{1}: \phi_{2}=5: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142203
Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is
1 $1: 4$
2 $1: 2$
3 $1: \sqrt{2}$
4 $1: 1$
Explanation:
C Maximum K.E = Incident photon energy work function $\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W}$ According to question- $\frac{1}{2} \mathrm{mv}_{1}^{2}=3 \mathrm{~W}-\mathrm{W}=2 \mathrm{~W}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=5 \mathrm{~W}-\mathrm{W}=4 \mathrm{~W}$ Dividing equation (i) by equation (ii), we get- $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{2}$
MHT-CET 2020
Dual nature of radiation and Matter
142204
The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $\mathrm{X}$ axis respectively are
1 Planck's constant, threshold frequency
2 work function, maximum K.E.
3 Planck's constant, work function
4 Maximum K.E, threshold frequency
Explanation:
A From Einstein's photoelectric equation, (K.E $)_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ Equation of straight line, Here $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$ Here $\mathrm{m}=\mathrm{h}$ $\mathrm{y}$-intercept $\mathrm{c}=-\phi_{\mathrm{o}}$ So, the slope of graph and intercept on $\mathrm{x}$-axis shown Planck's constant and threshold frequency.
MHT-CET 2020
Dual nature of radiation and Matter
142205
The light of wavelength ' $\lambda$ '. Incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light,$h=$ Planck's constant, $m=$ mass of electron)
142201
Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is
1 $3: 2$
2 $2: 3$
3 $1: 3$
4 $1: 2$
Explanation:
C We know that kinetic energy in photoelectric effect, $\mathrm{K}_{\max }=\mathrm{E}-\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=2 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=10 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=9 \phi_{\mathrm{o}}$ Dividing equation (i) by equation (ii), we get - $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{9}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{3}$ Hence, the maximum velocity ratio - $\mathrm{v}_{1}: \mathrm{v}_{2}=1: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142202
Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm}$. What is the ratio of work functions for lights of wavelength ' $\lambda_{1}$ ' to ' $\lambda_{2}$ '?
1 $6: 1$
2 $3: 5$
3 $5: 3$
4 $1: 6$
Explanation:
C Given that, $\lambda_{1}=360 \mathrm{~nm}, \lambda_{2}=600 \mathrm{~nm}$ Work function - $\phi=\frac{h \mathrm{c}}{\lambda}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{600}{360}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{5}{3}$ $\phi_{1}: \phi_{2}=5: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142203
Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is
1 $1: 4$
2 $1: 2$
3 $1: \sqrt{2}$
4 $1: 1$
Explanation:
C Maximum K.E = Incident photon energy work function $\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W}$ According to question- $\frac{1}{2} \mathrm{mv}_{1}^{2}=3 \mathrm{~W}-\mathrm{W}=2 \mathrm{~W}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=5 \mathrm{~W}-\mathrm{W}=4 \mathrm{~W}$ Dividing equation (i) by equation (ii), we get- $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{2}$
MHT-CET 2020
Dual nature of radiation and Matter
142204
The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $\mathrm{X}$ axis respectively are
1 Planck's constant, threshold frequency
2 work function, maximum K.E.
3 Planck's constant, work function
4 Maximum K.E, threshold frequency
Explanation:
A From Einstein's photoelectric equation, (K.E $)_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ Equation of straight line, Here $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$ Here $\mathrm{m}=\mathrm{h}$ $\mathrm{y}$-intercept $\mathrm{c}=-\phi_{\mathrm{o}}$ So, the slope of graph and intercept on $\mathrm{x}$-axis shown Planck's constant and threshold frequency.
MHT-CET 2020
Dual nature of radiation and Matter
142205
The light of wavelength ' $\lambda$ '. Incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light,$h=$ Planck's constant, $m=$ mass of electron)
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Dual nature of radiation and Matter
142201
Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is
1 $3: 2$
2 $2: 3$
3 $1: 3$
4 $1: 2$
Explanation:
C We know that kinetic energy in photoelectric effect, $\mathrm{K}_{\max }=\mathrm{E}-\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=2 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=10 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=9 \phi_{\mathrm{o}}$ Dividing equation (i) by equation (ii), we get - $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{9}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{3}$ Hence, the maximum velocity ratio - $\mathrm{v}_{1}: \mathrm{v}_{2}=1: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142202
Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm}$. What is the ratio of work functions for lights of wavelength ' $\lambda_{1}$ ' to ' $\lambda_{2}$ '?
1 $6: 1$
2 $3: 5$
3 $5: 3$
4 $1: 6$
Explanation:
C Given that, $\lambda_{1}=360 \mathrm{~nm}, \lambda_{2}=600 \mathrm{~nm}$ Work function - $\phi=\frac{h \mathrm{c}}{\lambda}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{600}{360}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{5}{3}$ $\phi_{1}: \phi_{2}=5: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142203
Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is
1 $1: 4$
2 $1: 2$
3 $1: \sqrt{2}$
4 $1: 1$
Explanation:
C Maximum K.E = Incident photon energy work function $\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W}$ According to question- $\frac{1}{2} \mathrm{mv}_{1}^{2}=3 \mathrm{~W}-\mathrm{W}=2 \mathrm{~W}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=5 \mathrm{~W}-\mathrm{W}=4 \mathrm{~W}$ Dividing equation (i) by equation (ii), we get- $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{2}$
MHT-CET 2020
Dual nature of radiation and Matter
142204
The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $\mathrm{X}$ axis respectively are
1 Planck's constant, threshold frequency
2 work function, maximum K.E.
3 Planck's constant, work function
4 Maximum K.E, threshold frequency
Explanation:
A From Einstein's photoelectric equation, (K.E $)_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ Equation of straight line, Here $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$ Here $\mathrm{m}=\mathrm{h}$ $\mathrm{y}$-intercept $\mathrm{c}=-\phi_{\mathrm{o}}$ So, the slope of graph and intercept on $\mathrm{x}$-axis shown Planck's constant and threshold frequency.
MHT-CET 2020
Dual nature of radiation and Matter
142205
The light of wavelength ' $\lambda$ '. Incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light,$h=$ Planck's constant, $m=$ mass of electron)
142201
Two incident radiations having energies two times and ten times of the work function of a metal surface, produce photoelectric effect. The ratio of maximum velocities of emitted photo electrons respectively is
1 $3: 2$
2 $2: 3$
3 $1: 3$
4 $1: 2$
Explanation:
C We know that kinetic energy in photoelectric effect, $\mathrm{K}_{\max }=\mathrm{E}-\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{1}^{2}=2 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=\phi_{\mathrm{o}}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=10 \phi_{\mathrm{o}}-\phi_{\mathrm{o}}=9 \phi_{\mathrm{o}}$ Dividing equation (i) by equation (ii), we get - $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{9}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{3}$ Hence, the maximum velocity ratio - $\mathrm{v}_{1}: \mathrm{v}_{2}=1: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142202
Photoelectrons are emitted from a photosensitive surface for the light of wavelengths $\lambda_{1}=360 \mathrm{~nm}$ and $\lambda_{2}=600 \mathrm{~nm}$. What is the ratio of work functions for lights of wavelength ' $\lambda_{1}$ ' to ' $\lambda_{2}$ '?
1 $6: 1$
2 $3: 5$
3 $5: 3$
4 $1: 6$
Explanation:
C Given that, $\lambda_{1}=360 \mathrm{~nm}, \lambda_{2}=600 \mathrm{~nm}$ Work function - $\phi=\frac{h \mathrm{c}}{\lambda}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{\lambda_{2}}{\lambda_{1}}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{600}{360}$ $\frac{\phi_{1}}{\phi_{2}}=\frac{5}{3}$ $\phi_{1}: \phi_{2}=5: 3$
MHT-CET 2020
Dual nature of radiation and Matter
142203
Energy of the incident photon on the metal surface is ' $3 W$ ' and then ' $5 W$ ', where ' $W$ ' is he work function for that metal. The ratio of velocities of emitted photoelectrons is
1 $1: 4$
2 $1: 2$
3 $1: \sqrt{2}$
4 $1: 1$
Explanation:
C Maximum K.E = Incident photon energy work function $\mathrm{K}_{\max }=\mathrm{E}-\mathrm{W}$ According to question- $\frac{1}{2} \mathrm{mv}_{1}^{2}=3 \mathrm{~W}-\mathrm{W}=2 \mathrm{~W}$ $\frac{1}{2} \mathrm{mv}_{2}^{2}=5 \mathrm{~W}-\mathrm{W}=4 \mathrm{~W}$ Dividing equation (i) by equation (ii), we get- $\frac{\mathrm{v}_{1}^{2}}{\mathrm{v}_{2}^{2}}=\frac{1}{2}$ $\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\frac{1}{\sqrt{2}}$ $\mathrm{v}_{1}: \mathrm{v}_{2}=1: \sqrt{2}$
MHT-CET 2020
Dual nature of radiation and Matter
142204
The graph of kinetic energy against the frequency ( $v$ ) of incident light is as shown in the figure. The slope of the graph and intercept on $\mathrm{X}$ axis respectively are
1 Planck's constant, threshold frequency
2 work function, maximum K.E.
3 Planck's constant, work function
4 Maximum K.E, threshold frequency
Explanation:
A From Einstein's photoelectric equation, (K.E $)_{\text {max }}=\mathrm{h} v-\phi_{\mathrm{o}}$ Equation of straight line, Here $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$ Here $\mathrm{m}=\mathrm{h}$ $\mathrm{y}$-intercept $\mathrm{c}=-\phi_{\mathrm{o}}$ So, the slope of graph and intercept on $\mathrm{x}$-axis shown Planck's constant and threshold frequency.
MHT-CET 2020
Dual nature of radiation and Matter
142205
The light of wavelength ' $\lambda$ '. Incident on the surface of metal having work function $\phi$ emits the electrons. The maximum velocity of electrons emitted is $(c=$ velocity of light,$h=$ Planck's constant, $m=$ mass of electron)
