(B) Given: Wavelength $\lambda = 488 \;nm = 488 \times 10^{-9} \;m$, Stopping potential $V_o = 0.38 \;V$.The energy of the incident photon is given by

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(B) Given: Wavelength $\lambda = 488 \;nm = 488 \times 10^{-9} \;m$, Stopping potential $V_o = 0.38 \;V$.The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $h = 6.63 \times 10^{-34} \;J \cdot s$ and $c = 3 \times 10^8 \;m/s$:$E = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{488 \times 10^{-9}} \approx 4.076 \times 10^{-19} \;J$.Converting to $eV$: $E = \frac{4.076 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.547 \;eV$.According to Einstein's photoelectric equation: $K_{max} = E - \phi_o$, where $K_{max} = eV_o$.Thus, $\phi_o = E - eV_o$.$\phi_o = 2.547 \;eV - 0.38 \;eV = 2.167 \;eV$.Rounding to the nearest option, the work function is $2.16 \;eV$.

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

Medium

Light of wavelength $488 \;nm$ is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is $0.38 \;V$. Find the work function (in $eV$) of the material from which the emitter is made.

A
$4.32$
B
$2.16$
C
$8.64$
D
$6.48$

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Dual Nature of Radiation and matter

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Einstein's Photoelectric Equation and Energy Quantum of Radiation

When radiation of wavelength $\lambda$ is incident on a metallic surface,the stopping potential of ejected photoelectrons is $4.8 \, V$. If the same surface is illuminated by radiation of double the previous wavelength,then the stopping potential becomes $1.6 \, V$. The threshold wavelength of the metal is $... \lambda$.

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An electron moving with velocity $1.6 \times 10^7 \ m/s$ has a wavelength of $0.4 \ \mathring{A}$. The required accelerating voltage for the electron motion is (charge on electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$).

MediumMHT CET 2023

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$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

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The maximum velocity of an electron emitted by light of wavelength $\lambda$ incident on the surface of a metal of work function $\phi$ is [$h=$ Planck's constant,$m=$ mass of electron and $c=$ speed of light]

MediumAP EAMCET 2020

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In a photoelectric experiment,ultraviolet light of wavelength $280 \, nm$ is used with a lithium cathode having a work function $\phi = 2.5 \, eV$. If the wavelength of incident light is switched to $400 \, nm$,find out the change in the stopping potential (in $V$).
$(h = 6.63 \times 10^{-34} \, J \cdot s, c = 3 \times 10^8 \, m/s)$

DifficultJEE MAIN 2021

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An electron moving with velocity $1.6 \times 10^7 \ m/s$ has a wavelength of $0.4 \ \mathring{A}$. The required accelerating voltage for the electron motion is (charge on electron $= 1.6 \times 10^{-19} \ C$,mass of electron $= 9 \times 10^{-31} \ kg$).

$A$ photon of energy $8 \ eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} \ Hz$. The maximum kinetic energy of the photoelectrons emitted (in $eV$) is: (Take $h = 6 \times 10^{-34} \ J \cdot s$ and $1 \ eV = 1.6 \times 10^{-19} \ J$)

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