Modern Physics 1 Question 58

54. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975\AA. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as 13.6eV.

(1982,5M)

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Answer:

Correct Answer: 54. Six, 1.875μm

Solution:

  1. Energy corresponding to given wavelength

E( in eV)=12375λ( in \AA)=12375975=12.69eV

Now, let the electron excites to nth  energy state. Then,

EnE1=12.69 or (13.6)(n2)(13.6)=12.69n4

i.e. electron excites to 4 th energy state. Total number of lines in emission spectrum would be

n(n1)2=4×32=6

Longest wavelength will correspond to the minimum energy and minimum energy is released in transition from n=4 to n=3.

E43=E4E3=13.6(42)13.6(3)2=0.66eV

Longest wavelength will be,

λmax=12375E( in eV)=123750.66\AA=1.875×106m=1.875μm



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