Answer :
Explanation:
It is given that,
Initial state of electron, [tex]n_i=1[/tex]
Final state of electron, [tex]n_f=3[/tex]
The wavelength of the excited electron is given by :
[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]
Where
R is Rydberg's constant
[tex]\dfrac{1}{\lambda}=1.097\times 10^{7}\ J\times (\dfrac{1}{3^2}-\dfrac{1}{1^2})[/tex]
[tex]\lambda=-1.02\times 10^{-7}\ m[/tex]
or
[tex]\lambda=102\ nm[/tex]
Let f is the frequency of the observed photon. It is given by :
[tex]f=\dfrac{c}{\lambda}[/tex]
[tex]f=\dfrac{3\times 10^8\ m/s}{1.02\times 10^{-7}\ m}[/tex]
[tex]f=2.94\times 10^{15}\ Hz[/tex]
Hence, this is the required solution.
Answer:
1.032 x 10^-7 m, 2.9 x 10^15 Hz
Explanation:
n = 1 to n = 3
Rydberg's constant, R = 1.09 × 10^7 per metre
Use the formula for the wavelength
[tex]\frac{1}{\lambda }=R\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )[/tex]
here, n1 = 1 and n2 = 3
[tex]\frac{1}{\lambda }=1.09\times 10^{7}\left ( \frac{1}{1^{2}}-\frac{1}{3^{2}} \right )[/tex]
[tex]\frac{1}{\lambda }=1.09\times 10^{7}\times \frac{8}{9}[/tex]
λ = 1.032 x 10^-7 m
Let the frequency is f.
Use the relation
v = f x λ
[tex]3 \times 10^8=f\times 1.032 \times 10^{-7}[/tex]
f = 2.9 x 10^15 Hz