Solved Numerical Problem No. 19.4 ( Unit-19: Dawn of Modern Physics); HSSC-II (Class-12); Physics (Punjab Curriculum and Text Board, Lahore) With Verified Answers

Yellow light 577nm wavelength is incident on a cesium surface. The stopping voltage is found to be 0.25 V. Find

(a) the maximum K.E. of the photoelectrons

and (b) the work function of cesium. (Answer: (a) 4 ｘ 10⁻²⁰ J (b) 1.9 eV )

Given data:

Wavelength of yellow light = λ =  577 nm = 577 ｘ 10⁻⁹ m

 

Stoppage Voltage = V₀ = 0.25 V

∴ Plank's Constant = h =  6.626 ｘ 10⁻³⁴ m² kg s⁻¹

∴ Velocity of light = c =  3 ｘ 10⁸ m s⁻¹

To Find:

(a) Maximum energy of the photo-electron = K.E =  ?

(b) Work function = ф = ?

Solution:

(a) Maximum energy of the photo-electron = K.E =  ?

The formula for finding the maximum energy of an electron according to given data is

`\K.E_{max}` =  V₀ e

by putting values

`\K.E_{max}` =  0.25 V ｘ 1.6 ｘ 10⁻¹⁹ C

`\K.E_{max}` = 0.4 ｘ 10⁻¹⁹ J

or

`\K.E_{max}` = 4 ｘ 10⁻²⁰ J -----------Ans.1

This is the maximum energy by Photo-electron.

(b) Work function = ф = ?

Using Einstein's photoelectric equation to find the maximum energy of the electron

`\K.E_{max}` = hf - ф 

or

 

ф = hf - `\K.E_{max}`

but Frequency f is unknown, so we have

V = f λ

or 

c = f λ      [ v = c ]  

or

f = `\frac {c}{λ}` 

So, equation (1) becomes

ф = h`\frac {c}{λ}` - `\K.E_{max}` ----------(2)

by putting values

ф = 6.626 ｘ 10⁻³⁴ m² kg s⁻¹ ｘ `\frac {3 ｘ 10⁸ m s⁻¹}{577 ｘ 10⁻⁹ m}` - 4 ｘ 10⁻²⁰ J

ф = 3.45 ｘ 10⁻¹⁹ J  - 0.4 ｘ 10⁻¹⁹ J

ф = 3.05 ｘ 10⁻¹⁹ J 

by expressing in electron Volts (eV) we have [ ∴ 1 eV = 1.6 ｘ 10⁻¹⁹ J ] So, using the unitary method we have

`\K.E_{max}` = `\frac {3.05 ｘ 10⁻¹⁹ }{1.6 ｘ 10⁻¹⁹ }` eV

`\K.E_{max}` =1.9 eV -------------- Ans.2

Similar Questions:

A metal, whose work function is 3.0 eV, is illuminated by light of wavelength 3 ｘ 10⁻⁷ m. Calculate the (a) threshold frequency, (b) maximum energy of photo-electrons and (c) stopping potential (Answer: (a) 0.72 ｘ 10¹⁵ Hz, (b) 1.16 eV (c) 1.16V)

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