Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron pep_e to that of photon pPhp_{Ph}, pepPh\frac{p_e}{p_{Ph}} is:

[Take speed of light = 3×1083 \times 10^8 ms−1^{-1}, charge of electron = −1.6×10−19-1.6 \times 10^{-19} C and mass of electron = 9×10−319 \times 10^{-31} kg]

Show answer and solution
Ee=EPh=20E_e = E_{Ph} = 20 eV

⇒Pe=2meEe,PPh=EPhc\Rightarrow P_e = \sqrt{2m_e E_e}, P_{Ph} = \frac{E_{Ph}}{c}

⇒PePPh=2meEeEPhc=2meEPhc\Rightarrow \frac{P_e}{P_{Ph}} = \frac{\sqrt{2m_e E_e}}{E_{Ph}} c = \sqrt{\frac{2m_e}{E_{Ph}}}c

PePPh=2×9×10−3120×1.6×10−19×3×108\frac{P_e}{P_{Ph}} = \sqrt{\frac{2 \times 9 \times 10^{-31}}{20 \times 1.6 \times 10^{-19}}} \times 3 \times 10^8

=34×10−6×3×108=9004=225= \frac{3}{4} \times 10^{-6} \times 3 \times 10^8 = \frac{900}{4} = 225

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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