JEE Main 26 Feb 2021 Shift 2 Solved Paper

As we know, mass of photon, m=1.6×10−27kg Using de-Broglie wavelength,
λ=‌

h

p

=‌

h

mv

. . . (i)
where, h is Planck's constant =6.63×10−34J−s and p is momentum of photon.
By using Rydberg wavelength equation,
‌

1

λ

‌‌=R[‌

1

n22

−‌

1

n12

]
‌

1

λ

‌‌=1.09×107[‌

1

12

−‌

1

25

]
⇒‌‌‌

1

λ

‌‌=1.09×107[‌

24

25

]
⇒‌‌λ‌‌=‌

25

1.09×107×24

Substituting values in Eq. (i), we get
v‌‌=‌

h

mλ

=‌

6.63×10−34×1.09×107×24

1.6×10−27×25

‌‌=‌

6.63×1.09×24

1.6×2.5

=4.17ms−1