(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $\nu_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9})

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(C) The frequency of the $H_{\alpha}$ line of the Balmer series is given by $\nu_0 = Rc(\frac{1}{2^2} - \frac{1}{3^2}) = Rc(\frac{1}{4} - \frac{1}{9}) = \frac{5Rc}{36}$. This is the threshold frequency $\nu_0$ of the material.The frequency of the $H_{\beta}$ line of the Balmer series is given by $\nu = Rc(\frac{1}{2^2} - \frac{1}{4^2}) = Rc(\frac{1}{4} - \frac{1}{16}) = \frac{3Rc}{16}$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{max}$ is given by $K_{max} = h\nu - h\nu_0$.Substituting the values:$K_{max} = h(\frac{3Rc}{16}) - h(\frac{5Rc}{36})$$K_{max} = Rhc(\frac{3}{16} - \frac{5}{36})$$K_{max} = Rhc(\frac{27 - 20}{144}) = \frac{7 Rhc}{144}$.

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

Medium

The threshold frequency of a photosensitive material is equal to the frequency of the $H_{\alpha}$ line of hydrogen. If a photon whose frequency is equal to the frequency of the $H_{\beta}$ line of hydrogen is incident on this photosensitive material,the maximum kinetic energy of the emitted photoelectrons is ($R$ = Rydberg's constant,$h$ = Planck's constant,and $c$ = speed of light in vacuum).

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A
$Rhc$
B
$\frac{5 Rhc}{144}$
C
$\frac{7 Rhc}{144}$
D
$\frac{Rhc}{36}$

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Class 12

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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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Assertion : When ultraviolet light is incident on a photocell,its stopping potential is $V_0$ and the maximum kinetic energy of the photoelectrons is $K_{max}$. When the ultraviolet light is replaced by $X-$ rays,both $V_0$ and $K_{max}$ increase.
Reason : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light.

EasyAIIMS 2013

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The work functions for the following metals are given:
$Na: 2.75\;eV; K: 2.30\;eV; Mo: 4.17\;eV; Ni: 5.15\;eV$. Which of these metals will not give photoelectric emission for a radiation of wavelength $3300\,\mathring{A}$ from a $He-Cd$ laser placed $1\;m$ away from the photocell? What happens if the laser is brought nearer and placed $50\;cm$ away?

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When a photon of wavelength $3000 \ \mathring{A}$ is incident on a metal surface,the energy of the emitted electron is $0.5 \ eV$. If a photon of wavelength $2000 \ \mathring{A}$ is incident on the same metal surface,the energy of the emitted electron will be:

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$A$ metal surface having work function '$W_{0}$' emits photoelectrons when photons of energy '$E$' are incident on it. The electron enters a uniform magnetic field '$B$' in a perpendicular direction and moves in a circular path of radius '$r$'. Then '$r$' is equal to (where '$m$' and '$e$' are the mass and charge of the electron,respectively).

DifficultMHT CET 2020

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The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

MediumMHT CET 2020

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When a photon of wavelength $3000 \ \mathring{A}$ is incident on a metal surface,the energy of the emitted electron is $0.5 \ eV$. If a photon of wavelength $2000 \ \mathring{A}$ is incident on the same metal surface,the energy of the emitted electron will be:

The threshold wavelengths for photoelectric emission from two metals $A$ and $B$ are $400 \ nm$ and $800 \ nm$ respectively. The ratio of their work functions,$\phi_{A} : \phi_{B}$ is:

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