the energy of an electron in bohr orbit of H atom is -13.6eV..the possible energy values of excited states?

Question

for elecrons in bohr orbits of hydrogen is ..?

Answer 1

-13.6/(n^2)

n can be any positive integer.

for n=1, it is just -13.6/(1^2) = -13.6

for n=2, it is -13.6/(2^2) = -13.6/4 which is approximately -3.4

for n=3... you get the idea

the units are always eV.

Answer 2

Energy level is the sum of kinetic energy and potential energy of an electron orbiting round a nucleus in a particular stable orbit.
E(n)= k.e +p.e
= 1/2 mv^2 + Qq/4(22/7)E.r
where Q and q is charges, E. =permittivity of free space, r= radius of atom
Qq/4(22/7)E.r = mv^2/r
mv^2= Ze^2/ [4(22/7)E.r]
E(n)= Ze^2/ [8(22/7)E.r] + (Ze)(-e)/ [4(22/7)E.r]
= -Ze^2/ 8(22/7)E.r
substitute r=n^2 (h^2 E.)/ [m(22/7)Ze^2]
for H atom, Z(mass number) = 1
m=mass of electron, h= Planck's constant

Energy values of certain energy level can be found by substituting n=1, 2, 3...

Exitation energy is that energy needed to promote an electron from its ground state to a higher energy level.
For H atom, the 1st exitation energy is 10.2eV.

Answer 3

It is easy if you know..

The energy of the electron in Bohr orbit with number n


is E = -13.6/ n^2

This holds if you do not take in account the number of subshell l.

In practice, the energy depends also of the subshell.

But in a first approximation, you can use that formula