Question

Calculate the energy(J) change associated with an electron transition from n=2 to n=5 in a Bohr hydrogen atom. Put answer in scientific notation.

Answer 1

Answer:

4.58*10^(-19) J

Explanation:

Using the Rydberg's equation:

$\frac{1}{\lambda} = R \times (\frac{1}{n^2_{final}} - \frac{1}{n^2_{initial}})$

where

$\lambda$ is  the wavelength of the photon;

R is the Rydberg's constant = 1.0974*10^7 m^(-1)

final level is 5 and initial level is 2.

$\frac{1}{\lambda} = 1.0974 \times 10^7 \times (\frac{1}{5^2} - \frac{1}{2^2})$

$\frac{1}{\lambda} = -2304540 m^{-1}$

$\lambda = -4.339 \times 10^{-7} \;m$

Energy change is calculated with the next formula:

E = h*c/λ

where h is the Planck's constant = 6.626*10^(-34) J*s, and c is the speed of light = 299,792,458 m/s

E = 6.626*10^(-34)*299,792,458/-4.339*10^(-7)

E = 4.58*10^(-19) J

Answer 2

The energy at n level of hydrogen atom energy level =13.6/n^2

substiture the respective n values in the equation above and find the difference in the energy levels

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