Question

The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV. What is the kinetic energy of the electron in this state?​

Answer 1

EXPLANATION.

Total energy of an electron in the first excited

state of the hydrogen atom = -3.4eV.

To find the kinetic energy of the electron

in this state.

According to the question,

$\rm \to \: in \: \: the \: \: bhor \: \: model$

$\rm \to \: mvr \: = nh$

$\rm \to \: \frac{mv {}^{2} }{r} = \frac{z {e}^{2} }{4\pi \epsilon. {r}^{2} }$

$\rm \to \: t \: = \frac{1}{2} mv {}^{2}$

$\rm \to \: \frac{1}{2} mv {}^{2} = \frac{z {e}^{2} }{8\pi \epsilon.r}$

$\rm \to \: r \: = \frac{4\pi \epsilon. {h}^{2} }{2 {e}^{2}m } \times {n}^{2}$

zero potential energy at infinity.

$\rm \to \: v \: = \frac{ - ze {}^{2} }{4\pi \epsilon.r}$

$\rm \to \: v \: = - 2t$

$\rm \to \: e \: = t \: + \: v$

$\rm \to \: e \: = \: - t$

Kinetic energy of the electron is equal to the

negative of the total energy.

$\rm \to \: k \: = - e$

$\rm \to \: k \: = - ( - 3.4)ev$

$\rm \to \: k \: = + 3.4ev$

Therefore,

The kinetic energy of the electron = +3.4eV.

Answer 2

Answer:

K.E of electron = 3.4 eV

Step-by-step explanation:

Given,

The total energy of an electron in the first excited state of the hydrogen atom is about −3.4 eV.

And, we know, The kinetic energy equals the negative of the total energy of an electron.

So, K.E of electron =  - ( - 3.4 eV) = 3.4 eV.

Therefore, the kinetic energy of the electron in this state is 3.4 eV.