if threshold wavelength for ejection of electron is 330 nm then what is the work function for the photoelectric emission? (in joules)
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threshold wavelength = λ₀ = 330nm
work function = W₀ = hv₀ = hc/λ₀ = 6.626 x 10³⁴ x 3 x 10⁸ / 330 x 10⁻⁹
= 19.878 x 10⁴² / 330 x 10⁻⁹ = 6.023 x 10 ⁴⁹ J
hope my solution is correct.
work function = W₀ = hv₀ = hc/λ₀ = 6.626 x 10³⁴ x 3 x 10⁸ / 330 x 10⁻⁹
= 19.878 x 10⁴² / 330 x 10⁻⁹ = 6.023 x 10 ⁴⁹ J
hope my solution is correct.
RohanJ:
sorry but answer is wrong, you used 10^34 instead of 10^-34 i found the correct answer it is 6.0236×10^-19J
thx by the way
Oh, I'm sorry.
don't be sorry
thx again
You're welcome. You can ask more doubts in chemistry to me.
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