#### Question

A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

*λ*_{1} = 3650 Å, *λ*_{2}= 4047 Å, *λ*_{3}= 4358 Å, *λ*_{4}= 5461 Å, *λ*_{5}= 6907 Å,

The stopping voltages, respectively, were measured to be:

*V*_{01} = 1.28 V, *V*_{02} = 0.95 V, *V*_{03} = 0.74 V, *V*_{04} = 0.16 V, *V*_{05} = 0 V

Determine the value of Planck’s constant *h*, the threshold frequency and work function for the material.

[*Note: *You will notice that to get *h *from the data, you will need to know *e *(which you can take to be 1.6 × 10^{−19} C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of *e *(from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of *h*.]

#### Solution

Einstein’s photoelectric equation is given as:

*eV*_{0} = *h**ν*− `phi_0`

`V_0 = h/e v - phi_0/e` ...(1)

Where,

*V*_{0} = Stopping potential

*h* = Planck’s constant

*e* = Charge on an electron

ν = Frequency of radiation

`phi_0` = Work function of a material

It can be concluded from equation (1) that potential *V*_{0} is directly proportional to frequency *ν*.

Frequency is also given by the relation:

`v = "Speed of light (c)"/"Wavelenght (λ)"`

This relation can be used to obtain the frequencies of the various lines of the given wavelengths.

`v_1 = c/lambda_1 = (3xx 10^8)/(3650 xx 10^(-10)) = 8.219 xx 10^14 Hz`

`v_2 = c/lambda_2 = (3xx10^8)/(4047 xx 10^(-10)) = 7.412 xx 10^14 Hz`

`v_3 = c/lambda_3 = (3xx10^8)/(4358 xx 10^(-10)) = 6.884 xx 10^14 Hz`

`v_4 = c/lambda_4 = (3 xx 10^8)/(5461 xx 10^(-10)) = 5.493 xx 10^14 Hz`

`v_5 = c/lambda_5 = (3xx10^8)/(6907 xx 10^(-10)) = 4.343 xx 10^14 Hz`

The given quantities can be listed in tabular form as:

0.16

Frequency × 10^{14} Hz |
8.219 |
7.412 |
6.884 |
5.493 | 4.343 |

Stopping potential V_{0} |
1.28 | 0.95 |
0.74 |
0.16 | 0 |

The following figure shows a graph between *ν*and *V*_{0.}

_{}

It can be observed that the obtained curve is a straight line. It intersects the *ν*-axis at 5 × 10^{14} Hz, which is the threshold frequency (*ν*0) of the material. Point D corresponds to a frequency less than the threshold frequency. Hence, there is no photoelectric emission for the *λ*5 line, and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = "AB"/"CB" = `(1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

From equation (1), the slope `h/e` can be written as:

`h/e = (1.28 - 0.16)/((8.214 - 5.493) xx 10^14)`

`:. h = (1.12 xx 1.6 xx 10^(-19))/(2.726 xx 10^(14))`

`= 6.573 xx 10^(-34) Js`

The work function of the metal is given as:

`phi_0` = *h*ν_{0}

= 6.573 × 10^{−34} × 5 × 10^{14}

= 3.286 × 10^{−19} J

`=(3.286 xx 10^(-19))/(1.6 xx 1^(-18))`= 2.054 eV