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Physics Question 40 – NEET-UG 2026

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Consider that an electron is revolving in an excited state of Hydrogen atom with velocity 25.6×105 ms−1. The radius of the orbit is x×10−9 m. The value of x is : [Take the mass of electron to be 9×10−31 kg, charge of electron =−1.6×10−19 C and 14πϵ0=9×109 N m2 C−2]

For an electron revolving in a stable orbit around a nucleus, the electrostatic force of attraction between the electron and the nucleus provides the necessary centripetal force.

Step 1: Identify the forces acting on the electron✦ Active

The electron revolves around the nucleus due to the electrostatic force of attraction, which provides the necessary centripetal force. For a hydrogen atom, the nucleus has a charge of +e.

Step 2: Equate the centripetal force and electrostatic force○ Expand

The centripetal force is Fc=mev2r and the electrostatic force is Fe=ke2r2. Equating them:

mev2r=ke2r2

This simplifies to mev2=ke2r, or r=ke2mev2.

💡 Teacher's Secret Hint

Remember that for a hydrogen atom, the nuclear charge is Z=1, so the electrostatic force is between e and e.

Step 3: Substitute the given values and calculate x○ Expand

Given k=9×109 N m2 C−2, e=1.6×10−19 C, me=9×10−31 kg, and v=25.6×105 ms−1 (so v2=25.6×1010 m2s−2). Substituting these values into the equation for r:

r=(9×109)×(1.6×10−19)2(9×10−31)×(25.6×1010)

Calculate the terms:

r=9×109×(2.56×10−38)9×25.6×10−21

Cancel out 9 and simplify the powers of 10:

r=2.56×109−3825.6×10−21=2.56×10−2925.6×10−21

Further simplification:

r=0.1×10−8=1×10−1×10−8=1×10−9 m

Comparing this with the given radius r=x×10−9 m, we find x=1.

💡 Teacher's Secret Hint

Pay close attention to the powers of 10 during calculation to avoid errors.

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