JEE Main 24 Feb 2021 Shift 2 Solved Paper

Given, V=1.24 million volt =1.24×106 volt
Since, energy (E)=eV
where, e is the charge of electron =1.6×10−19C
∴‌‌E=1.6×10−19×1.24×106 ... (i)
As we know that,
Energy of photon, E=‌

hc

λ

...(ii)
Here, Planck's constant, h=6.67×10−34J−s,
C= speed of light in free space, C=3×108ms−1
Equating Eqs. (i) and (ii), we get
1.6×10−19×1.24×106‌‌=‌

6.67×10−34×3×108

λ

⇒‌‌λ‌‌=‌

20.01×10−13

1.6×1.24

=10.09×10−13
‌‌=1.009×10−12∼eq10−3×10−9
‌‌=10−3nm