Question

A hydrogen atom in state n = 6 makes two successive transitions and reaches the ground state. In the first transition a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition (b) What is the value of n in the intermediate state?

Answer 1

Answer:

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Answer 2

Value of n is 3 in Intermediate State

Explanation:

• Energy at n = 6,  $E = \frac {-13.6}{36} = - 0.3777777$

• Energy in ground state = –13.6 eV  

• Energy emitted in Second transition = –13.6 –(0.37777 + 1.13)  

= –12.09 = 12.1 eV  

• b) Energy in the intermediate state = 1.13 ev + 0.0377777

= $1.507777 = \frac {13.6 \times Z^2}{n^2}$

= $\frac {13.6}{n^2}$

or, $n = \sqrt \frac {13.6}{1.507}$

=  3.03  

= 3 = n