The maximum velocity of the photoelectrons em | vedclass

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by $KE_{\max} = h n - \phi$,where $h$ is Planck's c

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more than $\sqrt{3} v$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $KE_{\max}$ is given by $KE_{\max} = h n - \phi$,where $h$ is Planck's constant,$n$ is the frequency,and $\phi$ is the work function of the metal.For the first case: $\frac{1}{2} m v^2 = h n - \phi$  ..... $(i)$For the second case with frequency $3n$: $\frac{1}{2} m (v^{\prime})^2 = 3 h n - \phi$  ..... $(ii)$From equation $(i)$,we have $h n = \frac{1}{2} m v^2 + \phi$. Substituting this into equation $(ii)$:$\frac{1}{2} m (v^{\prime})^2 = 3 (\frac{1}{2} m v^2 + \phi) - \phi$$\frac{1}{2} m (v^{\prime})^2 = \frac{3}{2} m v^2 + 2 \phi$$(v^{\prime})^2 = 3 v^2 + \frac{4 \phi}{m}$Since $\phi > 0$,it follows that $(v^{\prime})^2 > 3 v^2$,which implies $v^{\prime} > \sqrt{3} v$.Therefore,the maximum velocity will be more than $\sqrt{3} v$.

The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$,the maximum velocity of the ejected photoelectrons will be:

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