Question

If the momentum of an electron is increased by $P_m$ then the de Broglie wavelength associated with it changes by $0.5$%. Then the initial momentum of the electron is

• A. $100P_m$
• B. $\frac{P_m}{100}$
• C. $200P_m$
• D. $\frac{P_m}{200}$

Answer 1

The correct option is C $200P_m$

The De broglie wavelength $\lambda$ and momentum $p$ of a particle are related by ,

$p=\frac{h}{\lambda }$

as $p\propto 1/\lambda$ , therefore , with increase in momentum , wavelength will decrease .

Initially , $p=\frac{h}{\lambda }$ ................eq1

Now the wavelength is decreased by 0.5% ,

${\lambda }^{\prime }=\lambda -\lambda \times 0.5/100=199\lambda /200$

hence , increased momentum ,

$p+p_m=\frac{h}{{\lambda }^{\prime }}$

or $p+p_m=\frac{200h}{199\lambda }$ ............eq2

dividing eq2 by eq1 , we get

$\frac{p+p_m}{p}=\frac{200}{199}$

or $199p+200p_m=200p$

or $p=200p_m$