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Physics Question on de broglie hypothesis

The de Broglie wavelength of a proton (charge = 1.6×10−19C1.6\times10^{-19}C mass = 1.6×10−27kg)1.6\times10^{-27} kg) accelerated through a p.d of 1 kV1\, kV is

A

0.9 nm0.9\, nm

B

7 ?7\,?

C

0.9×10−12 m0.9 \times10^{-12}\, m

D

600 ?600\, ?

Answer

0.9×10−12 m0.9 \times10^{-12}\, m

Explanation

Solution

According to de-Broglie hypothesis
λ=hp\lambda = \frac{h}{p}
=h2mE=h2mqV= \frac{h}{\sqrt{2 m E}} = \frac{h}{\sqrt{2 mq V}}
∴λ=6.6×10−342×(1.6×10−27)(1.6×10−19)×1000\therefore \lambda = \frac{6.6 \times10^{-34}}{\sqrt{2 \times\left(1.6 \times10^{-27}\right) \left(1.6 \times10^{-19}\right) \times1000} }
=6.6×10−347.16×10−22= \frac{6.6 \times10^{-34}}{7.16 \times10^{-22}}
=0.9×10−12 m= 0.9 \times10^{-12} \,m