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Physics Question on de broglie hypothesis

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 (Given h=4.14×10−15eVs h = 4.14 \times 10^{-15} eVs and c=3×108 ms−1c = 3 \times 10^8 \, ms^{-1})

A

≈0.6×106 ms−1\approx 0.6 \times 10^6 \, ms^{-1}

B

≈61×103 ms−1\approx 61 \times 10^3 \, ms^{-1}

C

≈0.3×106 ms−1\approx 0.3 \times 10^6 \, ms^{-1}

D

≈6×105 ms−1\approx 6 \times 10^5 \, ms^{-1}

Answer

≈6×105 ms−1\approx 6 \times 10^5 \, ms^{-1}

Explanation

Solution

λ0=3250×10−10m\lambda_{0}=3250 \times 10^{-10} m
λ=2536×10−10m\lambda=2536 \times 10^{-10} m
ϕ=1242eV−nm325nm=3.82 eV\phi=\frac{1242 eV - nm }{325 nm }=3.82\, eV
hv=1242eV−nm253.6nm=4.89 eVh v=\frac{1242 eV - nm }{253.6 nm }=4.89\, eV
KEmax⁡=(4.89−3.82) eV=1.077 eVKE _{\max }=(4.89-3.82)\, eV =1.077\, eV
12mv2=1.077×1.6×10−19\frac{1}{2} m v^{2}=1.077 \times 1.6 \times 10^{-19}
v=2×1.077×1.6×10−199.1×10−31v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}}
v=0.6×106m/sv=0.6 \times 10^{6} m / s