Question

What is the energy of the photon released when an electron drops from n=7 to n=5 in a single electron atom?

4.27 x 10^{-20} J

1.98 x 10^{-19} J |

9.18 x 10^{-20} J |

5.01 x 10^{-19} J |

2.14 x 10^{-18} J |

Answer #1

Here photon will be emitted

1/wavelength = R* (1/nf^2 - 1/ni^2)

R is Rydberg constant. R = 1.097*10^7

1/wavelength = R* (1/nf^2 - 1/ni^2)

1/wavelength = 1.097*10^7* (1/5^2 - 1/7^2)

wavelength = 4.653*10^-6 m

wavelength = 4653 nm

we have:

wavelength = 4.653*10^-6 m

we have below equation to be used:

Energy = Planck constant*speed of light/wavelength

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.653*10^-6 m)

= 4.272*10^-20 J

Since this is emitted energy, sign will be negative

So,

E = 4.272*10^-20 J

Answer: 4.27*10^-20 J

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