Question: Q. 4. (i) Using Bohr’s second postulate of quantization of orbital angular momentum, show that the circumference of the electron in the nth  orbital state in hydrogen atom is n times the de-Broglie wavelength associated with it.

(ii) The electron in hydrogen atom is initially in the third excited state. What is the maximum number of spectral lines which can be emitted when it finally moves to the ground state?

U A [Delhi I, II, III 2012]

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Solution:

Ans. (i) According to Bohr’s second postulate

mvrn=nh2π 2πrn=nhmv

 But hmv=hp=λ 2πrn=nλ

[Note : If the student just writes mvrn=nh2π and writes λ=hp award 2 marks]

(ii) For third excited state, n=4

For ground state, n=1

Hence, possible transitions are

nf=4 to ni=3,2,1 nf=3 to ni=2,1 nf=2 to ni=1

Total number of transitions =6

Note : If transition is taking place nth orbit to 1st orbit, number of possible spectral lines are n(n1)2.

[CBSE Marking Scheme 2012]

Commonly Made Error

  • Some students are unable to recall the correct equations

 i.e., mvrn=nh2π and λ=hp



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