Answer
- When $n_i=3, \lambda=656nm$. This is the red line near the right end of the spectra line.
- When $n_i=4, \lambda=486nm$. This is the blue-green line near the $500nm$ point.
- When $n_i=5, \lambda=433.9nm$. This is the blue-violet line near the $450nm$ point.
Work Step by Step
*When $n_f=2$, we have:
$\Delta E= E_f - E_i=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{n^2_i})$
1) $n_i=3$:
$\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{9})\approx-3.028\times10^{-19}J$
So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is:
$E_{photon}=|\Delta E|\approx3.028\times10^{-19}J$
The wavelength of light emitted is:
$\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{3.028\times10^{-19}}\approx6.56\times10^{-7}m\approx656nm$
This is the red line near the right end of the line spectra.
2) $n_i=4$:
$\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{16})=-4.0875\times10^{-19}J$
So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is:
$E_{photon}=|\Delta E|=4.0875\times10^{-19}J$
The wavelength of light emitted is:
$\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.0875\times10^{-19}}\approx4.86\times10^{-7}m\approx486nm$
This is the blue-green line near the point $500nm$.
3) $n_i=5$:
$\Delta E=(-2.18\times10^{-18})\times(\frac{1}{4}-\frac{1}{25})=-4.578\times10^{-19}J$
So, the energy of a photon of light emitted when electron moves from n = 3 to n = 2 is:
$E_{photon}=|\Delta E|=4.578\times10^{-19}J$
The wavelength of light emitted is:
$\lambda=\frac{h\times c}{E_{photon}}=\frac{(6.626\times10^{-34})\times(2.998\times10^8)}{4.578\times10^{-19}}\approx4.339\times10^{-7}m\approx433.9nm$
This is the blue-violet line near the $450nm$ point.
