Question

The ratio of de-Broglie wavelengths of proton and $\alpha$-particle having same kinetic energy is

• A. $\sqrt{2}:1$
• B. $2\sqrt{2}:1$
• C. 2 :1
• D. 4 : 1

Answer 1

The correct option is C 2 :1

De Broglie wavelength is given by:

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

Writing momentum as a a function of kinetic energy and mass

$\lambda =\frac{h}{\sqrt{2Em}}$

i.e. $\frac{{\lambda }_{proton}}{{\lambda }_{alpha}}=\sqrt{\frac{m_{alpha}}{m_{proton}}}$

$m_{alpha}=4m_{proton}$
So,

$\frac{{\lambda }_{proton}}{{\lambda }_{alpha}}=\sqrt{\frac{m_{alpha}}{m_{proton}}}=\sqrt{\frac{4}{1}}=2$