1. ≈ 0.6×10⁶ ms ⁻¹ 2. ≈ 61×10³ ms ⁻¹ 3. ≈ 0.3×10⁶ ms ⁻¹ 4. ≈ 6×10⁵ ms ⁻¹

Answer: Dropped in the official answer key: full marks to every candidate Key idea: 1/2mv²=hc/λ-hc/λ₀ . hc=4.14×10⁻¹⁵×3×10⁸=1242 eV nm. Photon energy =1242/253.6=4.90 eV; work function =1242/325=3.82 eV. KE ≈1.08 eV =1.7×10⁻¹⁹ J. v=sqrtdfrac2×1.7×10⁻¹⁹9.1×10⁻³¹≈6×10⁵ m/s. Options 1 and 4 both equal this value, so the question has two correct options. The official key treats it as a bonus question.

Topic: Einstein's photoelectric equation, particle nature of light· Dual Nature of Matter and Radiation

The photoelectric threshold wavelength of silver is 3250×10^-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10^-10 m is :(Given h = 4.14×10^-15 eVs and c = 3×10^8 ms^-1)

1. \(\approx 0.6\times10^{6}\) ms\(^{-1}\)

2. \(\approx 61\times10^{3}\) ms\(^{-1}\)

3. \(\approx 0.3\times10^{6}\) ms\(^{-1}\)

4. \(\approx 6\times10^{5}\) ms\(^{-1}\)

Doubtly Asked 30 Oct 2026 NEET UG 2017 neet Save WhatsApp Report

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Answer: Dropped in the official answer key: full marks to every candidate

Key idea: \(\tfrac12mv^2=\dfrac{hc}{\lambda}-\dfrac{hc}{\lambda_0}\).

\(hc=4.14\times10^{-15}\times3\times10^8=1242\) eV nm.

Photon energy \(=\dfrac{1242}{253.6}=4.90\) eV; work function \(=\dfrac{1242}{325}=3.82\) eV. KE \(\approx1.08\) eV \(=1.7\times10^{-19}\) J.

\(v=\sqrt{\dfrac{2\times1.7\times10^{-19}}{9.1\times10^{-31}}}\approx6\times10^{5}\) m/s.

Options 1 and 4 both equal this value, so the question has two correct options. The official key treats it as a bonus question.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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The photoelectric threshold wavelength of silver is 3250×10^-10 m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10^-10 m is :(Given h = 4.14×10^-15 eVs and c = 3×10^8 ms^-1)

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