Question

Find the frequency of light emitted in the transition from the 148th orbit to the 139th orbit. Find the frequency of light absorbed in the transition from the 165th orbit to the 174th orbit. Submit Answer

Answer 1

Answer:

$f=3\times 10^8\times 64.90=19.4\times 10^9 s^{-1}$

$f=3\times 10^8\times39.9537=119.8613\times 10^8\ /sec$

Explanation:

for the transition from 148th orbit to 138th orbit

we know the formula

$\frac{1}{\lambda }=R\times \left ( \frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}} \right )$

where R is Rydberg constant whose value is  $1.0974\times 10^7 /m$

putting this value in equation

$\frac{1}{\lambda }=1.0974\times 10^7 \left ( \frac{1}{139^2}-\frac{1}{148^2} \right )$

$\frac{1}{\lambda }=64.9075$

$f=c\lambda$

$f=3\times 10^8\times 64.90=19.4\times 10^9 s^{-1}$

for the transition from 165th to  174th orbit

we know the formula

$\frac{1}{\lambda }=R\times \left ( \frac{1}{n^2_{final}}-\frac{1}{n^2_{initial}} \right )$

where R is Rydberg constant whose value is  $1.0974\times 10^7 /m$

putting this value in equation

$\frac{1}{\lambda }=1.0974\times 10^7 \left ( \frac{1}{165^2}-\frac{1}{174^2} \right )$

$\frac{1}{\lambda }=39.95377$

$f=c\lambda$

$f=3\times 10^8\times39.9537=119.8613\times 10^8\ /sec$