Q: A proton and an electron are accelerated by same potential difference have de-Broglie wavelength λ_{p} and λ_{e}

(A) λ_{e} = λ_{p}

(B) λ_{e} < λ_{p}

(C) λ_{e} > λ_{p}

(D) none of these

**Solution : **

As Proton & electron have same charge and accelerated by same Potential difference V . Therefore Kinetic Energy K = q V will be constant .

De-broglie Wavelength $\large \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m K}}$

$\large \lambda \propto \frac{1}{\sqrt{m}}$

$\large \frac{\lambda_p}{\lambda_e} = \sqrt{\frac{m_e}{m_p} }$

$\large \frac{\lambda_p}{\lambda_e} < 1 $ (Since m_{p} > m_{e})

Hence , λ_{e} > λ_{p}

Correct option is (C)