Question

1. The energy of the electron in the lowest level of the hydrogen atom (n=1) is -2.179×10-18 J. What is the energy of the electron in level n=5? -8.716×10-20 J 2.The electron in a hydrogen atom moves from level n=6 to level n=4.

a) Is a photon emitted or absorbed?

b) What is the wavelength of the photon?

Answer #1

Energy of an electron in n^{th} orbital is given by,

E= -(2.179x10^{-18})/n^{2} Joules

, therefore, we can write as,

=

therefore,

E_{5} =
(-2.179x10^{-18})(1^{2}/5^{2})

E_{5} = -8.176x10^{-20}

a) as the electron energy is directly proportional to the square of the orbit number,

when the electron moves from n=6 to n=4,

its energy decreases, so it must have emitted a photon so that the energy will decrease.

b) E_{6}-E_{4} =
(E_{1}/n_{6}^{2}) -
(E_{1}/n_{4}^{2}) (From above
equations)

hence,

E_{6}-E_{4} = 7.5659 x 10^{-20}
Joules

now this energy difference will be equal to (hc/)

where h = Planck's constant

c = velocity of light

= wavelength of photon

= hc / change in energy

=((6.634x10^{-34})x(3x10^{8}))/7.5659 x
10^{-20}

=2.63x10^{-6} m

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