Question

An electron in a Hydrogen atom originally at n=5 energy level absorbs a photon with a frequency of 6.54×10^13 and then proceeds to emit another photon with a frequency of 2.98×10^14. To what energy level does the electron move?

Answer 1

Energy needed for transition from level n1 to n2 in hydrogen is given by bohr’s formula,

∆E=hv=(me^4/8eo^2c h^3) (1/n1^2-1/n2^2)………….eqn1

V=frequency absorbed

(me^4/8eo^2c h^3)=Rh/hc

Rh=109667.6 cm-1=rydberg’s constant

h=planck’s const,c=speed of light

in the given situation ,Net energy absorbed=hv1-hv2=h(v1-v2)=(6.54*10^13-2.98*10^14)s-1

Net energy absorbed=h*(6.54*10^13-2.98*10^14)s-1

                                  =-23.26*10^13*h

Putting in eqn 1,

∆E=hv= Rh/hc (1/n1^2-1/n2^2)

-23.26*10^13*h = Rh/hc (1/n1^2-1/n2^2)   

Or, -23.26*10^13 = Rh/c (1/n1^2-1/n2^2)   

Or, -23.26*10^13/c =109667.6 cm-1 (1/n1^2-1/n2^2)   

Or, , -23.26*10^13 s-1/3*10^10cm/s =109667.6 cm-1 (1/5^2-1/n2^2)   

Or,7.7*10^3=109667.6 cm-1 (1/n2^2-1/25)

Or,7.02*10^-2= (1/n2^2-1/25)

7.02*10^-2 +0.04= 1/n2^2

0.1102=1/n2^2

n2^2=1/0.1102=9.07

n2=(9.07)^1/2=3

n2=3 it will go to level 3