JEE PYQ: Atomic Physics Question 66
Question 66 - 2019 (12 Jan Shift 2)
A particle A of mass ’m’ and charge ‘q’ is accelerated by a potential difference of 50 V. Another particle B of mass ‘4m’ and charge ‘q’ is accelerated by a potential difference of 2500 V. The ratio of de-Broglie wavelength $\frac{\lambda_A}{\lambda_B}$ is:
(1) 10.00
(2) 0.07
(3) 14.14
(4) 4.47
Show Answer
Answer: (3)
Solution
de Broglie wavelength $\lambda$ is given by $K = qV$. $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$ ($\because p = \sqrt{2mK}$). $\frac{\lambda_A}{\lambda_B} = \frac{\sqrt{2m_B q_B V_B}}{\sqrt{2m_A q_A V_A}} = \sqrt{\frac{4m \cdot q \cdot 2500}{m \cdot q \cdot 50}} = 2\sqrt{50} = 2 \times 7.07 = 14.14$.
BETA
