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  4. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 mathringA is 0.42 V. If the threshold frequency is x × 1013 / s, where x is (nearest integer). (Given, speed light =3 × 108 m / s, Planck's constant =6.63 × 10-34 Js )

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Q. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength $6630\,\mathring{A}$ is $0.42\, V$. If the threshold frequency is $x \times$ $10^{13} / s$, where $x$ is _____ (nearest integer).
(Given, speed light $=3 \times 10^{8} m / s$, Planck's constant $=6.63 \times 10^{-34} \, Js$ )

JEE MainJEE Main 2022Dual Nature of Radiation and Matter

Solution:

Stopping potential $V _{0}=0.42 V$
$\lambda=6630 \,\mathring{A}$
$ E =\phi+ eV _{0}$
$E$ : energy of incident photon
$V _{0}$ : Stopping potential
$\phi = E - eV _{0} $
$E =\frac{12400}{6630} eV =1.87 eV$
$\phi =(1.87-0.42)=1.45 eV $
$\phi = hv _{0} ; v _{0}: $threshold frequency
$1.45 \times 1.6 \times 10^{-19}=6.63 \times 10^{-34} \times v _{0}$
$v _{0} =0.35 \times 10^{15} $
$=35 \times 10^{13} \sec ^{-1}$
$=35$