JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Modern Physics

The kinetic energy of an electron,

α

-particle and a proton are given as

4 K

2 K

and K respectively. The de-Broglie wavelength associated with electron (

λ e

)

α

-particle (

λ α

) and the proton (

λ p

) are as follows:

(A) $λ α = λ p < λ e$
(B) $λ α > λ p > λ e$
(C) $λ α < λ p < λ e$ ✓ Correct
(D) $λ α = λ p > λ e$

Answer: (C)

Solution

The de-Broglie wavelength is

λ = \frac{h}{2 m K}

.

For the electron (mass

m /1840

, KE =

4 K

λ_{e} = \frac{h}{2 \cdot \frac{m}{1840} \cdot 4 K} = \frac{h}{\frac{8 m K}{1840}}

For the proton (mass

m

, KE =

K

λ_{p} = \frac{h}{2 m K}

For the

α

-particle (mass

4 m

, KE =

2 K

λ_{α} = \frac{h}{2 \cdot 4 m \cdot 2 K} = \frac{h}{16 m K}

Comparing denominators:

16 m K > 2 m K > \frac{8 m K}{1840}

Therefore:

λ_{α} < λ_{p} < λ_{e}

JEE Main · 2023 · 06 Apr 2023 · Shift 1 · Single correct option

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