First slide

Hydrogen -like ion

Question

A hydrogen like atom of atomic number Z is in excited state of quantum number 3n. The maximum energy which can be emitted by atom is 404 eV. If it makes the transition to quantum state n, a photon energy 40.4 eV is emitted. The value of n will be

Difficult

A

2
B

3
C

4
D

5

Solution

$i n t h e f i r s t c a s e m a x i m u m e n e r g y i s a t r a n s i t i o n f r o m 3 n t o 1 e n e r g y r e l e a s e d 13.6 x Z^{2} (\frac{1}{1^{2}} - \frac{1}{9 n^{2}}) = 404 - - - (1) a n d i n t h e s e c o n d c a s e a t r a n s i t i o n f r o m 3 n t o n 13.6 x Z^{2} (\frac{1}{n^{2}} - \frac{1}{9 n^{2}}) = 40.4 - - - (2) d i v i d i n g w e g e t \frac{9 n^{2} - 1}{8} = 10 n^{2} = 9 n = 3$

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