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Question

(D) The frequency of emitted radiation is given by $
u = \frac{\Delta E}{h}$.Emission occurs when an electron jumps from a higher energy level to a l

Vedclass · Accepted Answer

$n = 5 	o n = 2$

Vedclass · Answer

(D) The frequency of emitted radiation is given by $
u = \frac{\Delta E}{h}$.Emission occurs when an electron jumps from a higher energy level to a lower energy level.The energy difference $\Delta E$ is proportional to the frequency $
u$.For emission,we look for transitions from higher to lower levels ($n=3 	o n=1$ or $n=5 	o n=2$).The energy difference for $n=5 	o n=2$ is $\Delta E = 13.6 	imes Z^2 	imes (\frac{1}{2^2} - \frac{1}{5^2}) = 13.6 	imes (0.25 - 0.04) = 13.6 	imes 0.21 = 2.856 \ eV$.The energy difference for $n=3 	o n=1$ is $\Delta E = 13.6 	imes (1 - \frac{1}{9}) = 13.6 	imes 0.888 = 12.08 \ eV$.Since $2.856 \ eV < 12.08 \ eV$,the transition $n=5 	o n=2$ has the lowest energy change and therefore the lowest frequency.

In which transition of the hydrogen atom will the emitted radiation have the lowest frequency?

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