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Question

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by: $K_{\max} = E - \phi$ wh

Vedclass · Accepted Answer

$\left[\frac{2(hc-\phi \lambda)}{m \lambda}\right]^{\frac{1}{2}}$

Vedclass · Answer

(D) According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max}$ of the emitted electron is given by: $K_{\max} = E - \phi$ where $E = \frac{hc}{\lambda}$ is the energy of the incident photon. Substituting the values,we get: $\frac{1}{2} mv_{\max}^2 = \frac{hc}{\lambda} - \phi$ $\frac{1}{2} mv_{\max}^2 = \frac{hc - \phi \lambda}{\lambda}$ Multiplying both sides by $\frac{2}{m}$: $v_{\max}^2 = \frac{2(hc - \phi \lambda)}{m \lambda}$ Taking the square root on both sides: $v_{\max} = \sqrt{\frac{2(hc - \phi \lambda)}{m \lambda}}$ Thus,the correct option is $D$.

The light of wavelength $\lambda$ is incident on the surface of a metal with work function $\phi$ and emits electrons. What is the maximum velocity of the emitted electrons? [Given: $m=$ mass of electron,$h=$ Planck's constant,$c=$ velocity of light]

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