Find the velocity of an electron emitted by a metal whose threshold frequency is 2.46×1014 s−1...

Find the velocity of an electron emitted by a metal whose threshold frequency is 2.46×10¹⁴ s−1 when it is exposed to visible light of wavelength 4.79×10⁻⁷ m .

Expert Solution

E = hc/

= 6.625*10^-34 * 3*10^8/4.79*10^-7

= 4.15*10^-19 J

threshold frequency is 2.46×10¹⁴ s−1

E = hv₀

= 6.625*10^-34*2.46*10^14 = 1.63*10^-19 J

The emitted electron energy ( Kinetic energy) = energy of photon - threshold energy

= 4.15*10^-19- 1.63*10^-19 = (4.15-1.63)*10^-19 = 2.52*10^-19 J

1/2 mv^2 = 2.52*10^-19

mv^2 = 2.52*10^-19 *2 = 5.04*10^-19

V^2 = 5.04*10^-19/9.1*10^-31

V^2 = 5.54*10^11

V = 7.44*10^5 m/sec >>>>answer

velocity of electron = 7.44*10^5 m/sec

queen_honey_blossom answered 2 years ago

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