In a hydrogen like atom, when |NUCLEAR PHYSICS|JEEMAIN 2019

JEE MAIN 2019

11-01-2019 S2

Question

In a hydrogen like atom, when an electron jumps from the M-shell to the L-shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from N -shell to the L -shell, the wavelength of emitted radiation will be:

Select the correct option:

$\frac{25}{16} \lambda$

$\frac{16}{25} \lambda$

$\frac{20}{27} \lambda$

$\frac{27}{20} \lambda$

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$$\frac{1}{\lambda} = RZ^2 \left[ \frac{1}{4} - \frac{1}{9} \right] = \frac{5RZ^2}{36}$$$$\frac{1}{\lambda_1} = RZ^2 \left[ \frac{1}{4} - \frac{1}{16} \right] = \frac{3RZ^2}{16}$$$$\lambda_1 = \frac{16}{3RZ^2}, \lambda = \frac{36}{5RZ^2}$$$$\frac{\lambda_1}{\lambda} = \frac{16 \times 5}{3 \times 36} = \frac{20}{27}$$$$\lambda_1 = \frac{20}{27}\lambda$$

Question Tags

JEE Main

Physics

Medium

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