Light of wavelength 500nm falls on a metal whose work function is 1.9 eV . Find (1) the energy of the photon in eV (2) the kinetic energy of the photoelectron emitted and (3) the stopping potential. Given h=6.625x10raise to -34 Js.

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Asked by rutujasarangmehta | 15 Mar, 2016, 06:36: PM

Expert Answer

λ = 500 nm = 5000 × 10^-10

begin mathsize 14px style straight E space equals space hv space equals space hc over straight lambda straight E equals fraction numerator 6.63 cross times 10 to the power of negative 34 end exponent cross times 3 cross times 10 to the power of 8 over denominator 5000 cross times 10 to the power of negative 10 end exponent end fraction equals 3.978 cross times 10 to the power of negative 19 end exponent end style

1 eV = 1.6 × 10^-19 J

So,

begin mathsize 14px style 1 space straight J space equals space fraction numerator 1 over denominator 1.6 cross times 10 to the power of negative 19 end exponent eV end fraction end style

begin mathsize 14px style therefore 3.978 cross times 10 to the power of negative 19 end exponent space straight J space equals space fraction numerator 1 cross times 3.978 cross times 10 to the power of negative 19 end exponent over denominator 1.6 cross times 10 to the power of negative 19 end exponent end fraction equals 2.486 eV end style

Hence, the energy of the photon is 2.486eV.

E = KE + hv₀

'E' is the energy of incident photon, KE is the kinetic energy of emitted photon, hv₀ is the work function or minimum energy required.

Work function is 1.9eV.

KE = E - hv₀

= 2.486 - 1.9 = 0.586eV

Thus the kinetic energy of the photoelectron emitted is 0.586eV.

Answered by Faiza Lambe | 16 Mar, 2016, 01:25: PM

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