JEE Main 17 March 2021 Shift 1 Solved Paper

We know
λ=‌

h

p

...(i)
where, λ= wavelength,
h= Planck's constant
and ‌‌p= linear momentum
AlsO, kinetic energy (KE)=‌

1

2

mv2=‌

p2

2m

⇒‌‌p=√2mKE ...(ii)
From Eqs. (i) and (ii), we get
λ=‌

h

√2mKE

or λ=‌

h

√2mE

‌‌[∵KE=E]
Wavelength of electron, λe=‌

h

√2mE

and wavelength of photon, λp=‌

hc

E

[∵ Both electron and photon have same energy. ]
⇒‌‌‌

λe

λp

=‌

‌ h √2mE

hc E

⇒‌

λe

λp

=‌

1

c

(‌

E

2m

)1/2