BITSAT 2012 Paper

Given φ=2.25eV
λ=300×100−3m
Energy of the given photon, E=hv
=‌

hc

2

=‌

663×10−34×3×108

300×10−9

=6.63×10−19J=4.13eV
Now, E=‌

hc

2

−φ=4.137−2.25=1.88eV→(1)
So, 1.88eV energy is used to jump from one orbit to another orbit by electron. Therefore, energy of different orbital of hydrogen ⇒n=1‌‌2‌‌3‌‌4
E=−13.6ev&−3.4&−1.51&−0.85
And from statement (1), 1.18=Ei−Ef
{Ei&Ef are energy of initial and final orbit\}
And also from the above table, we can observe that, Ei−Ef=(−1.51)−(−3.4)=1.89eV
=E2−E3
That the electron jumps from 3rd to 2nd orbit. Therefore, with one photon of 300nm wavelength hydrogen electron can jump from 3 to 2 .