Question

Hey frnd plss tell me some numerical with solution of hydrogen spectrum ??? its urgent plsss ☺☺☺☺

Answer 1

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Answer 2

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Q1 ) Calculate the wave number of shortet wavelength transition in the Balmer series of atomic hydrogen??

Q2) calculate the wavelength of radiation which excites the electron of hydrogen atom from ground state to fourth energy level. Ionization energy of hydrogen atom is 1312 Kj/mol

Q3) the hydrogen atom in the ground state is excited by means of monochromatic radiation of wavelength x A° . The resulting spectrum consists of 15 different lines. Calculate the value of x.

Q5) The longest wave length transition in Lyman series of atomic hydrogen will be?

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Answers :

1) For Balmer series,

n1 = 2

Shortest wavelength means highest energy transition

So, n2 = ∞

1/λ = 1.09*10^7 (1/2^2 - 1/∞^2)

λ = 1.09*10^-7 * 1/4

λ = 2.725 * 10^-6

2) Ground state to Fourth Energy state :

n1 = 2 and n2 = 4

1/λ = 1.09*10^7 (1/1 - 1/16)

1/λ = 1.09*10^7(15/16)

λ = 16/1.09*15*10^7

λ = 0.978 * 10^-7

λ = 978A°

3) Here n1 = 1 and n2 = 2

1/λ = 1.09 * 10^7 (1 - 1/4)

1/λ = 1.09*3*10^7/4

λ= 1.2156*10^-7

λ = 1216 A°

4) Here n1 = 1

and

Since the number of different lines is 15, the excited state must be n = 6. As only then the number of lines in Lyman = 5, Balmer = 4, Paschen = 3, Brackett = 2, Pfund = 1. Sum of those different lines = 5 + 4 + 3 + 2 + 1 = 15.

So,

1/λ = 1.09*10^7 (1 - 1/36)

1/λ = 1.09*10^7 * 35 /36

λ = 36/1.09*35*10^7

λ = 0.943 * 10^-7

λ = 943 A°