Question

The debroglie wavelength of an electron moving with velocity v = 3c/5 is

Answer 1

Answer:

3.8 pM

Explanation:

Answer 2

Answer:

The De-Broglie wavelength of the electron is $0.036A^{0}$

Explanation:

Given that,

The velocity of the moving electron is $v=\frac{3c}{5}$

Here,c is the velocity of light which is $c=3\times10^{8}m/s$ hence our velocity becomes

$v=\frac{3}{5} \times 3\times10^{8}=1.8\times10^{8}m/s$

According to De-broglie,The Debroglie wavelength of an moving electron in an orbit is the ratio of planck`s constant to its momentum.Therefore,

$\lambda=\frac{h}{p}= \frac{h}{mv}$

The mass of an electron is $m=9.1\times10^{-31}kg$ and

The planck`s constant is $h=6.626\times10^{-34}Js$

Substituting the known values in De-broglie wavelength relation,we get

$\lambda=\frac{6.626\times10^{-34}}{(9.1\times10^{-31})(1.8\times10^{8})} =3.6\times10^{-12}m=0.036A^{0}$

Therefore,the De-Broglie wavelength of the electron is $0.036A^{0}$

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