Question

A proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
Given: $h = 6.63 \times 10^{-34} \, \text{Js}$, $m_e = 9.0 \times 10^{-31} \, \text{kg}$, and $m_p = 1836 \times m_e$

• A. $1 : 1836$
• B. $1 : \frac{1}{1836}$
• C. $1 : \frac{1}{\sqrt{1836}}$
• D. $1 : \sqrt{1836}$

Answer 1

Answer: (A)

$1 : 1836$

Answer 2

For the same de-Broglie wavelength, $P = \frac{h}{\lambda}$ is the same for both the proton and the electron. Kinetic energy is given by:
$\text{KE} = \frac{P^2}{2m}.$
Thus:
$\frac{\text{KE}_e}{\text{KE}_p} = \frac{m_p}{m_e}.$
Given:
$m_p = 1836 m_e.$
Substitute:
$\frac{\text{KE}_e}{\text{KE}_p} = \frac{1}{1836}.$
Final Answer: $1 : 1836$.