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  • The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6A, The number of emission lines will be (according to Bohr’s theory )

The ionisation potential of H-atom is 13.6V. When it is excited from ground state by monochromatic radiations of 970.6A, The number of emission lines will be (according to Bohr’s theory )

Option: 1

10


Option: 2

8


Option: 3

6


Option: 4

4


Answers (1)

best_answer

\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right] \\

\frac{1}{970.6 \times 10^{-10}}=1.097 \times 10^7\left[\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{n}_2^2}\right] \Rightarrow \mathrm{n}_2=4

\therefore Number \: of\: emission\; line \mathrm{N}=\frac{\mathrm{n}(\mathrm{n}-1)}{2}=\frac{4 \times 3}{2}=6

Posted by

Sanket Gandhi

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