The electrostatic potential energy of an electron in a Bohr orbit is -3.02 eV . Its kinetic energy and total energy are, respectively,

Correct answer: B. 1.51 eV, \,-1.51 eV For an electron in a hydrogen-like atom, the potential energy (PE), kinetic energy (KE), and total energy (TE) are related by: TE = PE 2 and KE = -TE Given PE = -3.02 eV , TE = (-3.02)/(2) = -1.51 eV KE = -(-1.51) = 1.51 eV Therefore, the kinetic and total energies are 1.51 eV and -1.51 eV , respectively. Answer: 1.51 eV, \,-1.51 eV

NEET

Exam: NEET
Subject: Physics
Chapter: Atomic Physics

NEET · Physics · Atomic Physics

The electrostatic potential energy of an electron in a Bohr orbit...

The electrostatic potential energy of an electron in a Bohr orbit is

-3.02 \text{ eV}

. Its kinetic energy and total energy are, respectively,

Options

A $6.04 \text{ eV}, \,-6.04 \text{ eV}$
B $1.51 \text{ eV}, \,-1.51 \text{ eV}$
C $-1.51 \text{ eV}, \,1.51 \text{ eV}$
D $3.02 \text{ eV}, \,-3.02 \text{ eV}$

Correct answer

B. $1.51 \text{ eV}, \,-1.51 \text{ eV}$

Step-by-step solution

For an electron in a hydrogen-like atom, the potential energy (PE), kinetic energy (KE), and total energy (TE) are related by:

\text{TE} = \dfrac{\text{PE}}{2}

and

\text{KE} = -\text{TE}

Given

\text{PE} = -3.02 \text{ eV}

\text{TE} = \dfrac{-3.02}{2} = -1.51 \text{ eV}

\text{KE} = -(-1.51) = 1.51 \text{ eV}

Therefore, the kinetic and total energies are

1.51 \text{ eV}

and

-1.51 \text{ eV}

, respectively.

Answer:

1.51 \text{ eV}, \,-1.51 \text{ eV}

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