Question

How much energy is emitted when an electron jumps from n=6 to n=4? What is the wavelength of the emitted light? What is the momentum of the photon?

Answer 1

The energy emitted can be found as

E = Rhc ( 1/n1² - 1/n2²)

where R is rydberg's constant, h is planck's constant and c is speed of light

E = 1.0974e7 * 6.626e-34 * 3e8 ( 1 / 6² - 1/ 4²)

E = -7.574e-20 J

the negative sign tells us that energy was released and photons came out.

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Now, To find the wavelength

E = hc / $\lambda$

$\lambda$ = hc / E

$\lambda$ = 6.626e-34 * 3e8 / 7.574e-20

$\lambda$ = 2.625e-6 m

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Momentum of photon

p = h / $\lambda$

p = 2.52e-28 Kg.m/s