(C) According to Einstein's photoelectric equation,the stopping potential $V_{s}$ is given by: $eV_{s} = h\nu - \phi$ $V_{s} = \frac{h}{e}\nu - \frac{

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(C) According to Einstein's photoelectric equation,the stopping potential $V_{s}$ is given by: $eV_{s} = h\nu - \phi$ $V_{s} = \frac{h}{e}\nu - \frac{\phi}{e}$ Comparing this with the equation of a straight line $y = mx + c$,the x-intercept (where $V_{s} = 0$) is the threshold frequency $\nu_{0}$,where $\nu_{0} = \frac{\phi}{h}$ or $\phi = h\nu_{0}$. From the given graph,the threshold frequency for metal '$P$' is $\nu_{0}$ and for metal '$Q$' is $\nu_{0}^{\prime}$. Since $\nu_{0} Therefore,the work function $\phi_{P} < \phi_{Q}$.

Class 12 Physics Dual Nature of Radiation and matter Einstein's Photoelectric Equation and Energy Quantum of Radiation

Easy

The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

MHT CET 2020 All MHT CET

A
$\phi_{P} = \phi_{Q}$
B
$\nu_{0} < \nu_{0}^{\prime}$
C
$\phi_{P} < \phi_{Q}$
D
$\phi_{P} > \phi_{Q}$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

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When $UV$ light of wavelength $300 \, nm$ is incident on a metal surface having a work function of $2.13 \, eV$, electron emission takes place. The stopping potential is: (Given $hc = 1240 \, eV \cdot nm$) (in $ \, V$)

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The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

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The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

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If a photocell is illuminated with a radiation of $1240 \, Å$, the stopping potential is found to be $8 \, V$; then the work function of the emitter and the threshold wavelength are

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If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

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The graph of stopping potential $(V_{s})$ against frequency $(\nu)$ of incident radiation is plotted for two different metals '$P$' and '$Q$' as shown in the graph. If $\phi_{P}$ and $\phi_{Q}$ are the work functions of metals '$P$' and '$Q$' respectively,then which of the following is correct?

Similar Questions

When $UV$ light of wavelength $300 \, nm$ is incident on a metal surface having a work function of $2.13 \, eV$, electron emission takes place. The stopping potential is: (Given $hc = 1240 \, eV \cdot nm$) (in $ \, V$)

The energy of a photon with wavelength $\lambda$ is $2 \ eV$. When it strikes a metal surface,the maximum velocity of the emitted photoelectrons is $v$. If the value of $\lambda$ is decreased by $25\%$ and the maximum velocity is doubled,the work function of the metal becomes ...... $eV$.

The figure represents the graph of photocurrent $I$ versus applied voltage $(V)$. The maximum kinetic energy of the emitted photoelectrons is:

If a photocell is illuminated with a radiation of $1240 \, Å$, the stopping potential is found to be $8 \, V$; then the work function of the emitter and the threshold wavelength are

If the work function for a certain metal is $3.2 \times 10^{-19} \ J$ and it is illuminated with light of frequency $8 \times 10^{14} \ Hz$,the maximum kinetic energy of the photoelectrons would be in joule:

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