calculate the frequency of light associated with the transition of an electron in a hydrogen atom from the n…

calculate the frequency of light associated with the transition of an electron in a hydrogen atom from the n = 2 to the n = 3 energy level.\no 1.79×10^{14} s^{-1}\no 5.59×10^{14} s^{-1}\no 3.28×10^{14} s^{-1}\no 4.57×10^{14} s^{-1}\no 2.19×10^{14} s^{-1}

calculate the frequency of light associated with the transition of an electron in a hydrogen atom from the n = 2 to the n = 3 energy level.\no 1.79×10^{14} s^{-1}\no 5.59×10^{14} s^{-1}\no 3.28×10^{14} s^{-1}\no 4.57×10^{14} s^{-1}\no 2.19×10^{14} s^{-1}

Answer

Explanation:

Step1: Recall the Rydberg - frequency formula

$v = R_Hc\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ where $R_H = 1.097\times 10^{7}\ m^{-1}$ is the Rydberg constant, $c= 3\times 10^{8}\ m/s$ is the speed of light, $n_1 = 2$ and $n_2 = 3$.

Step2: Substitute the values into the formula

$v=1.097\times 10^{7}\ m^{-1}\times3\times 10^{8}\ m/s\times\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ First, calculate $\frac{1}{2^2}-\frac{1}{3^2}=\frac{1}{4}-\frac{1}{9}=\frac{9 - 4}{36}=\frac{5}{36}$. Then, $v = 1.097\times 10^{7}\times3\times 10^{8}\times\frac{5}{36}\ s^{-1}$. $v=\frac{1.097\times3\times5\times 10^{7 + 8}}{36}\ s^{-1}=\frac{16.455\times 10^{15}}{36}\ s^{-1}\approx4.57\times 10^{14}\ s^{-1}$

Answer:

$4.57\times 10^{14}\ s^{-1}$