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NEET 2017
Exam
NEET
Year
2017
Subject
Physics
Chapter
Dual Nature of Matter

NEET 2017 · Physics · Dual Nature of Matter

The photoelectric threshold wavelength of silver is 3250 × 10 ^ –...

The photoelectric threshold wavelength of silver is 3250×10–10m3250 \times {10}^{– 10}m. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10–10m2536 \times {10}^{– 10}m is:-

(Givenh=4.14×10−15eVsandc=3×108ms−1)\left(Givenh = 4.14 \times {10}^{- 15}eVsandc = 3 \times {10}^{8}m{s}^{- 1}\right)

Options

  1. A≈6×106ms−1\approx 6 \times {10}^{6}m{s}^{- 1}
  2. B≈0.6×106ms−1\approx 0.6 \times {10}^{6}m{s}^{- 1}
  3. C≈61×103ms−1\approx 61 \times {10}^{3}m{s}^{- 1}
  4. D≈0.3×106ms−1\approx 0.3 \times {10}^{6}m{s}^{- 1}

Correct answer

B. ≈0.6×106ms−1\approx 0.6 \times {10}^{6}m{s}^{- 1}

Step-by-step solution

λ0=3250×10−10m{\lambda}_{0} = 3250 \times {10}^{- 10}m

λ=2536×10−10m\lambda = 2536 \times {10}^{- 10}m

ϕ=hcλ0=1242eV−nm325nm=3.82eVϕ = \dfrac{hc}{{\lambda}_{0}} = \dfrac{1242eV - \text{n}\text{m}}{325\text{n}\text{m}} = 3.82eV

hν=hcλ=1242eV−nm253.6nm=4.89eVhν = \dfrac{hc}{\lambda} = \dfrac{1242eV - \text{nm}}{253 . 6\text{nm}} = 4 .89 eV

By Einstein’s photoelectric equation

{K}_\max = \dfrac{hc}{\lambda} - hcλ0\dfrac{hc}{{\lambda}_{0}}

K{E}_\max = \left(4.89 - 3.82\right)eV = 1.07eV

12mv2=1.07×1.6×10−19\dfrac{1}{2}m{v}^{2} = 1.07 \times 1.6 \times {10}^{- 19}

v=2×1.07×1.6×10−199.1×10−31v = \sqrt{\dfrac{2 \times 1.07 \times 1.6 \times {10}^{- 19}}{9.1 \times {10}^{- 31}}}

v=0.6×106m/sv = 0.6 \times {10}^{6}m/s
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