JEE PYQ: Atomic Physics Question 3
Question 3 - 2021 (16 Mar Shift 2)
The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths?
($m_P = 1.00727$ u, $m_e = 0.00055$ u)
(1) 1860 : 1
(2) $(1860)^2$ : 1
(3) 41.4 : 1
(4) 43 : 1
Show Answer
Answer: (4)
Solution
$\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}$. $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2}{m_1}} = \sqrt{\frac{m_p}{m_e}} = \sqrt{1831.4} = 42.79 \approx 43$.
BETA
