Question

K.E of electron in a particular orbit is 3.4ev,
Potential energy is​

Answer 1

Answer:

The energy of an electron in Bohr’s orbit of hydrogen atom is given by the expression:



Since Z = 1 for hydrogen above equation can be further simplified to:

En = -13.6/n2 eV

Solution:

The energies of electrons in the Bohr's orbits of hydrogen atom expressed in eV are:

 Orbit  Energy 1 -13.6/12 = -13.6 eV2 -13.6/22 = -3.4 eV3 -13.6/32 = -1.51 eV4 -13.6/42 = -0.85 eV

Excited state(s) represent n = 2, 3, 4 ...... (greater than 1).

Note: 

The ratio of energy of electrons in the orbits of hydrogen atom is: 

E1 : E2 : E3 : E4 ........... = 1/12 : 1/22 : 1/32 : 1/42 .......... = 1 : 1/4 : 1/9 : 1/16 ..........

 

Conclusion:

Correct option is "a".

Explanation:

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