JEE Main 2022 · 27 Jun · Shift 1 · Physics Q43

PhysicsSingle correct+4 / -1Question 43 of 90

An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $(\lambda_\alpha:\lambda_{C12})$ is :

1:\sqrt3

\sqrt3:1

3 : 1

2:\sqrt3

Answer and worked solution

Official answer: B

Same KE. $\lambda=\dfrac h{\sqrt{2mK}}\propto\dfrac1{\sqrt m}$ : $\dfrac{\lambda_\alpha}{\lambda_C}=\sqrt{\dfrac{12}4}=\sqrt3$ .

(B)

Topics:de Broglie waves

JEE Main 2022 · 27 Jun · Shift 1, Physics question 43