1 Physics -- Photons

Explain why there are no photoelectrons emitted when the frequency of the incident light is less than 5.6 x 10¹⁴ Hz.(i) Calculate the work function for sodium. (ii)Use the graph to calculate the value of the Planck constant in Js.[Hint: Gradient =[ (1.82-0) x 1.6 x 10-¹⁹] / (4.4 x 10¹⁴)...

Explain why there are no photoelectrons emitted when the frequency of the incident light is less than 5.6 x 10¹⁴ Hz.

(i) Calculate the work function for sodium. (ii)Use the graph to calculate the value of the Planck constant in Js.

[Hint: Gradient =[ (1.82-0) x 1.6 x 10-¹⁹] / (4.4 x 10¹⁴) which leads , h = 6.6 x 10-³⁴ J s]

1 Answers

When the frequency of the incident light is less than 5.6 x 10¹⁴ Hz, no photoelectrons are emitted because the energy of the incident photons is less than the work function of sodium, and hence the electrons do not have enough energy to overcome the work function and be emitted.

(i) We can calculate the work function of sodium using the equation:

Energy of incident photon = work function + maximum kinetic energy of photoelectrons

At the threshold frequency, the maximum kinetic energy of photoelectrons is zero. Therefore, the energy of the incident photon is equal to the work function.

Energy of incident photon = hf

where h is Planck's constant and f is the frequency of the incident radiation.

The threshold frequency of sodium is 5.6 x 10¹⁴ Hz. Substituting this value into the above equation:

Energy of incident photon = hf = (6.626 x 10^-34 J s) x (5.6 x 10¹⁴ Hz)

= 3.72 x 10^-19 J

Therefore, the work function of sodium is 3.72 x 10^-19 J.

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Considering, 1 as high input/output and 0 as low input/output

OR Gate (Truth Table)

First Input(A)	Second Input(B)	Output(Y)
1	1	1
1	0	1
0	1	1
0	0	0

NAND Gate (Truth Table)

First Input(A)	Second Input(B)	A.B(AND output)	NAND Output(Y)
1	1	1	0
0	1	0	1
0	0	0	1
1	0	0	1

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