Question

If an electron and a proton have the same kinetic energy, the ratio of the de Broglie wavelengths of proton and electron would approximately be :

• A. 1 : 1837
• B. 43 : 1
• C. 1837 : 1
• D. 1 : 43

Answer 1

The correct option is D 1 : 43

$E=\frac{1}{2}m{v}^2$

$p=mv=\sqrt{2mE}$ $=\sqrt{2mqv}$

De Broglie wavelength associated with change particle

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2mE}}=\frac{h}{\sqrt{2mq\nu }}$

where $E=\frac{1}{2}m{v}^2=K.E,$ P is momentum of change

So, ${\lambda }_P=\frac{h}{\sqrt{2m_{p}E}}$

${\lambda }_e=\frac{h}{\sqrt{2m_{e}E}}$

So, $\frac{{\lambda }_p}{{\lambda }_e}=\frac{\sqrt{m_e}}{\sqrt{m_p}}$

$=\sqrt{\frac{m_e}{m_p}}$

$=\sqrt{\frac{9.1\times {10}^{-31}}{1.67\times {10}^{-27}}}$

$=\sqrt{5.449\times {10}^{-4}}$

$=2.33\times {10}^{-2}$

$=0.0233$

$=\frac{1}{43}$

So, the answer is option (D).