Question

The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is(a) – 6.5 eV(b) – 27.2 eV(c) 13.6 eV(d) – 13.6 eV

Answer 1

Total energy in the first orbit is -13.6eV

|Total energy| =kinetic energy

So

Kinetic energy=13.6eV

Answer 2

Answer:

The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is -13.6 ev.

(d) is correct option.

Explanation:

We know that,

The energy of an electron in Bohr's orbit of hydrogen atom  is

$E_{n}=-13.6\dfrac{z^2}{n^2}$

If $E_{n}$ is the total energy of an electron revolving is an orbit

Then, the kinetic energy is

$K.E =E_{n}$

Put the value of  $E_{n}$

$K.E=-13.6\dfrac{z^2}{n^2}\ eV$

For hydrogen, z = 1 and n = first Bohr orbit

The kinetic energy is

$K.E=-13.6\dfrac{1^2}{1^2}$

$K.E=-13.6\ eV$

Hence, The kinetic energy of electron in the first Bohr orbit of the hydrogen atom is -13.6 ev.