JEE PYQ: Atomic Physics Question 57
Question 57 - 2019 (12 Apr Shift 2)
Consider an electron in a hydrogen atom, revolving in its second excited state (having radius 4.65 A). The de-Broglie wavelength of this electron is:
(1) 3.5 A
(2) 6.6 A
(3) 12.9 A
(4) 9.7 A
Show Answer
Answer: (4)
Solution
$v = \frac{c}{137n} = \frac{c}{137 \times 3}$. $\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{m \times \frac{n \times c}{3 \times 137}} = \frac{h}{mc} \times (3 \times 137) = 9.7$ A.
BETA
