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The energy of a photon is equal to the KE of a proton. The energy of the photon is E. Let $ \lambda_1$ be the de- Broglie wavelength of the proton and $ \lambda_2$ be the wavelength of the photon. So, $\large\frac{\lambda_1}{\lambda_2}$ is

$\begin {array} {1 1} (a)\;E^0 & \quad (b)\;E^{\large\frac{1}{2}} \\ (c)\;E^{-1} & \quad (d)\;E^{-2} \end {array}$

 

1 Answer

Ans : (b)
$ \lambda = \large\frac{h}{p} \: \; \: \; \: $ photon energy $E = hv$
So, $ \lambda_1 = \large\frac{h}{p} = \large\frac{h}{ \sqrt {2mE}}$………..(i)
$ \Rightarrow E = hv_2 = \large\frac{hc}{ \lambda_2}$
$ \lambda_2 = \large\frac{hc}{E}$……………(ii)
From Eqns. (i) & (ii),
$ \large\frac{\lambda_1}{ \lambda_2} = \large\frac{h\sqrt {2mE}}{ \bigg( \large\frac{hc}{E} \bigg)} \sim \large\frac{E}{ \sqrt E} = E^{\large\frac{1}{2}}$

 

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

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