JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 18)
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:
$${{25} \over {16}}$$ $$\lambda $$
$${{27} \over {20}}$$ $$\lambda $$
$${{16} \over {25}}$$ $$\lambda $$
$${{20} \over {27}}$$ $$\lambda $$
Explanation
For M $$ \to $$ L steel
$${1 \over \lambda }$$ = K $$\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$$
for N $$ \to $$ L
$${1 \over {\lambda '}}$$ = K$$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$$
$$\lambda ' = {{20} \over {27}}\lambda $$
$${1 \over \lambda }$$ = K $$\left( {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right) = {{K \times 5} \over {36}}$$
for N $$ \to $$ L
$${1 \over {\lambda '}}$$ = K$$\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right) = {{K \times 3} \over {16}}$$
$$\lambda ' = {{20} \over {27}}\lambda $$
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