Calculate the energy and frequency of the rad | vedclass

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Question

(A) The energy difference for an electron transition in a hydrogen atom is given by $\Delta E = 2.18 	imes 10^{-18} \ J \left( \frac{1}{n_{1}^{2}} -

Vedclass · Accepted Answer

$E = 3.03 	imes 10^{-19} \ J, 
u = 4.57 	imes 10^{14} \ Hz$

Vedclass · Answer

(A) The energy difference for an electron transition in a hydrogen atom is given by $\Delta E = 2.18 	imes 10^{-18} \ J \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ight)$.For $n_{1} = 2$ and $n_{2} = 3$:$\Delta E = 2.18 	imes 10^{-18} \ J \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} ight) = 2.18 	imes 10^{-18} \left( \frac{1}{4} - \frac{1}{9} ight) = 2.18 	imes 10^{-18} \left( \frac{5}{36} ight) = 3.03 	imes 10^{-19} \ J$.The frequency $
u$ is calculated using $\Delta E = h
u$:$
u = \frac{\Delta E}{h} = \frac{3.03 	imes 10^{-19} \ J}{6.626 	imes 10^{-34} \ J \cdot s} = 4.57 	imes 10^{14} \ Hz$.

Calculate the energy and frequency of the radiation emitted when an electron jumps from $n = 3$ to $n = 2$ in a hydrogen atom.

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