Q 12-11-092JEE MainJEE Main 2023 (11 Apr, Shift 2)Easy
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy is (Assume $m_p=m_e\times1849$)
Answer: (A) $1:43$
$\lambda=\dfrac h{\sqrt{2mK}}\propto\dfrac1{\sqrt m}$: $\dfrac{\lambda_p}{\lambda_e}=\dfrac1{\sqrt{1849}}=\dfrac1{43}$.
Solution by Sreeraj P, M.Sc Physics