Q31 / 201 · NTA Abhyas (NEET) 2020 · Physics · Atomic Physics

De Broglie wavelength of $0.05 e V$ thermal neutron is

Correct answer
A
$1.3 Å$
B
$2 Å$
C
$5.4 Å$
D
$8 Å$

Solution

λ = \frac{h}{p} = \frac{h}{\sqrt{2 m_{0} K}}

λ = \frac{h}{\sqrt{2 \times 1 .66 \times 10^{- 27} \times 0 .05 \times 1 .6 \times 10^{- 19}}}

λ = 1 .28 Å

λ \approx 1 .3 Å

Hence, the correct answer is option: 'A'.

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