The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100 V is

Question

The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100 V is (A)  9.1 ×10-24  (B)  6.5 ×10-24  (C)  5.4 ×10-24  (D)  1.6 ×10-24

Tardigrade Admin · Accepted Answer

de-Broglie wavelength of electron is given by

λ = \frac{h}{m v} = \frac{h}{2 m E}

Substituting the value of

E

, we get

λ = \frac{h}{2 m e V}

Here

m = 9.1 \times 1 0^{- 31} k g; e = 1.6 \times 1 0^{- 19} C

and

h = 6.6 \times 1 0^{- 34} J s

we get

λ = \frac{12.27}{V} \times 1 0^{- 10} = \frac{12.27}{V} \overset{˚}{A}

The de-Broglie wavelength of electrons, when accelerated through a potential difference of

100 V

will be

λ = \frac{12.27}{100} = 1.227 \overset{˚}{A}

Moreover,

λ = \frac{h}{p}

\Rightarrow p = \frac{6.6 \times 1 0 ^{- 34}}{1.227 \times 1 0 ^{- 10}}

= 5.5 \times 1 0^{- 24} k g - m s^{- 1}

Q. The linear momentum of an electron, initially at rest, accelerated through a potential difference of $100 V$ is

$9.1 \times 1 0^{- 24}$

$6.5 \times 1 0^{- 24}$

$5.4 \times 1 0^{- 24}$

$1.6 \times 1 0^{- 24}$

Q. The linear momentum of an electron, initially at rest, accelerated through a potential difference of 100V is

9.1×10−24

6.5×10−24

5.4×10−24

1.6×10−24

Q. The linear momentum of an electron, initially at rest, accelerated through a potential difference of $100 V$ is

$9.1 \times 1 0^{- 24}$

$6.5 \times 1 0^{- 24}$

$5.4 \times 1 0^{- 24}$

$1.6 \times 1 0^{- 24}$