The wavelength of an electron of kinetic energy $4.50 \times 10^{-29} \mathrm{~J}$ is...... $\times 10^{-5} \mathrm{~m}$. (Nearest integer) Given: mass of electron is $9 \times 10^{-31} \mathrm{~kg}$, $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Atomic Structure

The wavelength of an electron of kinetic energy

4.50 \times 1 0^{- 29} J

is......

\times 1 0^{- 5} m

. (Nearest integer)

Given: mass of electron is

9 \times 1 0^{- 31} kg

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

Answer: 7

For a particle of kinetic energy

E

, momentum is

p = 2 m E

, so the de Broglie wavelength is

λ = \frac{h}{2 m E}

. Substituting

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

m = 9 \times 1 0^{- 31} kg

and

E = 4.5 \times 1 0^{- 29} J

gives the denominator

2 \times 9 \times 1 0^{- 31} \times 4.5 \times 1 0^{- 29} = 81 \times 1 0^{- 60} = 9 \times 1 0^{- 30}

. Therefore

λ = \frac{6.6 \times 1 0 ^{- 34}}{9 \times 1 0 ^{- 30}} = 0.733 \times 1 0^{- 4} = 7.33 \times 1 0^{- 5} m

. Expressed as a multiple of

1 0^{- 5} m

, the nearest integer value is 7.

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Integer answer