Question : A monochromatic light is incident on a hydrogen sample in ground state. Hydrogen atoms absorb a fraction of light and subsequently emit radiation of six different wavelengths. The frequency of incident light is x×1015Hz. The value of x is __________. [Given h=4.25×10-15 eVs]
Doubt by Ankit
Solution :
We know,
Maximum number of spectral line are given by the formula
[n(n-1)]/2
where n is the orbit from which the electron is falling back i.e. n2.
ATQ
[n(n-1)]/2 = 6
n(n-1) = 12
n2-n=12
n2-n-12=0
n2-4n+3n-12=0
n(n-4)+3(n-4)=0
(n-4)(n+3)=0
n-4=0 OR n+3=0
n=4 OR n=-3
but n can't be negative.
so n=-3 is rejected.
Hence, n=4 i.e. n2=4 and n1=1
E=E4-E1
hν=0.85-(-13.6)
hν=12.75
ν=12.75/4.25×10-15
v=3×1015 Hz
on comparing with
v=x×1015 Hz
Hence, we get x=3.