Question

How do you calculate the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level? Recall that for hydrogen En=−2.18×10^−18J(1/n^2)

Answer 1

Answer:

Wavelength, $\lambda=9.75\times 10^{-8}\ m$

Explanation:

We need to find the the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level. The energy is given by :

$\Delta E=E_1-E_4$

$\Delta E=-2.18\times 10^{-18}(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\Delta E=-2.04\times 10^{-18}\ J$

The energy of a photon is given by :

$E=\dfrac{hc}{\lambda}$

$\lambda=\dfrac{hc}{E}$

$\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.04\times 10^{-18}}$

$\lambda=9.75\times 10^{-8}\ m$

So, the wavelength of the light emitted by a hydrogen atom during a transition of its electron from the n = 4 to the n = 1 principal energy level is $\lambda=9.75\times 10^{-8}\ m$. Hence, this is the required solution.