Dual Nature Of Radiation And Matter Ques 6
6. The de-Broglie wavelength of neutron in thermal equilibrium at temperature $T$ is
[NEET Kar. 2013]
(a) $\frac{30.8}{\sqrt{T}} \AA$
(b) $\frac{3.08}{\sqrt{T}} \AA$
(c) $\frac{0.308}{\sqrt{T}} \AA$
(d) $\frac{0.0308}{\sqrt{T}} \AA$
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Answer:
Correct Answer: 6.(a)
Solution:
- (a) From formula $\lambda=\frac{h}{\sqrt{2 m K T}}$
$=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 1.67 \times 10^{-27} \times 1.38 \times 10^{-23} T}} m$
[By placing value of $h, m$ and $k$ ]
$=\frac{30.8}{\sqrt{T}} \AA$
BETA
