Question

A photon is absorbed by an electron that is in the n = 3 state of a hydrogen atom, causing the hydrogen atom to become ionized. Very far away from the nucleus, the released electron has a velocity of 750,000 m/s. What was the wavelength of the absorbed photon?

Answer 1

Answer:

$\lambda=3.99*10^{-7}m$

Explanation:

According to the law of conservation of energy, the energy of the absorbed photon must be equal to the binding energy of the electron plus the energy of the released electron:

$E_p=E_b+E_r\\\frac{hc}{\lambda}=\frac{13.6eV}{n^2}+\frac{m_ev^2}{2}$

1 eV is equal to $1.6*10^{-19}J$, so:

$13.6eV*\frac{1.6*10^{-19}J}{1eV}=2.18*10^{-18}J$

Solving for $\lambda$ and replacing the given values:

$\lambda=\frac{hc}{\frac{2.18*10^{-18}J}{n^2}+\frac{m_ev^2}{2}}\\\lambda=\frac{6.63*10^{-134}J\cdot s(3*10^8\frac{m}{s})}{\frac{2.18*10^{-18}J}{3^2}+\frac{(9.11*10^{-31}kg)(7.5*10^5\frac{m}{s})^2}{2}}\\\lambda=3.99*10^{-7}m$