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A photosensitive metallic surface has work function, $hv_{\circ}$. If photons of energy $2hv_{\circ}$ fall on this surface, the electrons come out with a maximum velocity of $4 \times 10^6 \: m/s$. When the photon energy is increased to $5hv_{\circ}$, then maximum velocity of photoelectrons will be

$\begin {array} {1 1} (a)\;2 \times 10^7\: m/s & \quad (b)\;2 \times 10^6\: m/s \\ (c)\;8 \times 10^6\: m/s & \quad (d)\;8 \times 10^5\: m/s \end {array}$

 

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Ans : (c)
$ \large\frac{1}{2} mv^2_{max} = hv - hv_{\circ}$
$ \large\frac{1}{2} m \times 4 \times 10^6 \times 4 \times 10^6 = 2v_{\circ} – hv_{\circ}$
Again, $ \large\frac{1}{2} m \times v^2_{max} = 5hv_{\circ} – hv_{\circ} $
Dividing, $\large\frac{v^2_{max} }{(4 \times 10^6 )^2} = \large\frac{4hv_{\circ}}{hv_{\circ}} = 4$
$ \Rightarrow \large\frac{v_{max}}{4 \times 10^6} = 2$
$ \Rightarrow v_{max}\: = 8 \times 10^6 m/s$

 

answered Dec 31, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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