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Question

(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of t

Vedclass · Accepted Answer

$\frac{hc}{2\lambda}$

Vedclass · Answer

(B) Let $\phi_{0}$ be the work function of the surface of the material. According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted photoelectrons in the first case is $K_{\max 1} = \frac{hc}{\lambda} - \phi_{0}$.In the second case,the wavelength is $\lambda/2$,so the maximum kinetic energy is $K_{\max 2} = \frac{hc}{\lambda/2} - \phi_{0} = \frac{2hc}{\lambda} - \phi_{0}$.Given that $K_{\max 2} = 3 K_{\max 1}$,we substitute the expressions:$\frac{2hc}{\lambda} - \phi_{0} = 3 \left( \frac{hc}{\lambda} - \phi_{0} \right)$.Expanding the equation: $\frac{2hc}{\lambda} - \phi_{0} = \frac{3hc}{\lambda} - 3\phi_{0}$.Rearranging the terms: $3\phi_{0} - \phi_{0} = \frac{3hc}{\lambda} - \frac{2hc}{\lambda}$.$2\phi_{0} = \frac{hc}{\lambda}$.Therefore,the work function is $\phi_{0} = \frac{hc}{2\lambda}$.

Similar Questions

When a photon with energy $8\ eV$ is incident on a metal surface having a threshold frequency of $1.6 \times 10^{15}\ Hz$,the maximum kinetic energy of the emitted photoelectrons is ............ $eV$. (Given: $h = 6.6 \times 10^{-34}\ J\cdot s$,$1\ eV = 1.6 \times 10^{-19}\ J$)

When a light of wavelength $\lambda$ falls on the emitter of a photocell,the maximum speed of emitted photoelectrons is $V$. If the incident wavelength is changed to $2\lambda / 3$,the maximum speed of the emitted photoelectrons will be:

$A$ 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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