JEE Main 31-Aug-2021 Shift 1 Solved Paper

Let de-Broglie wavelength of proton and electron are λp and λe.
Kinetic energy of electron and proton are Ke and Kp.
Momentum of electron and proton are pe and pP.
According to question, λp=λe=λ
By using de-Broglie wavelength,
λ=

h

p

=

h

√2‌mK

where, m is mass and h is Planck’s constant.
∴p=

h

λ

...(i)
From Eq. (i)
pe=pp as λe=λp
Again, K=

h2

2mλ

...(ii)
From Eq. (ii)
∴

Ke

Kp

=

mp

me

×

λp

λe

⇒

Ke

Kp

=

mp

me

[∵λe=λp]
∵mp>me
Ke>Kp