In a photoelectric effect experiment, visible light (400nm to 750nm) shines on a

Question

In a photoelectric effect experiment, visible light (400nm to 750nm) shines on a metal with a work function of 2.8eV.

1)What is the maximum KE of ejected electrons?

*a) 0.30 eV

            b) 1.30 eV

            c) 2.30 eV

            d) 3.30 eV

e) 4.30 eV

2)What is the cutoff frequency for this metal?

a) 4.77x1014 Hz

            b) 5.77x1014 Hz

*c) 6.77x1014 Hz

            d) 7.77x1014 Hz

e) 8.77x1014 Hz

3)Which of the following metals’ work functions would emit no electrons when visible light shines on it?

a) 0 eV

            b) 1.0 eV

            c) 2.0 eV

            d) 3.0 eV

*e) 4.0 eV

The answers are marked by * but would like an explanation on how these answers were obtained. Thanks in advance!

Explanation / Answer

1.

the maximum KE of ejected electrons:

KE = hf - W = hc /lambda - W = [1240 eV/400] - 2.8 eV = 0.3 eV

2

cutoff frequency for this metal:

fo = W/h = 2.8 eV/6.63x10-34 = 2.8*1.6x10-19/6.626x10-34 = 6.77e+14

3.

below the cuttof frequency there is are no electrons are emitted.

the work-functions 0 to 3 eV gives the frequency above the cuttoff frequency, but the 4 eV

workfunction gives the less value of cutttoff frequency. so, 4 eV is crrect option.

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