NCERT Class XI Chemistry Structure of Atom Solutions

Question : 51

Total: 67

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold frequency and (b) the threshold wavelength of the radiation. (c) If the caesium elements is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Solution:

(a) Work function = hυ0 = 1.9 eV = 1.9 × 1.602 × 10–19J = 3.04 × 10–19J
Threshold frequency, (u0) =

3.04×10−19

6.626×10−34

= 4.59 × 1014sec−1
(b) Threshold wavelength, (λ0) =

3×108

4.59×1014

= 5.64 × 10−7m
or 654 × 10–9m or 654 nm
(c) Now, energy of light, (E) =

6.626×10−34×3×108

500×10−9

= 3.98 × 10−19J
Kinetic energy of ejected electron = 3.98 × 10–19 – 3.04 × 10–19 = 9.4 × 10–20J
Since K.E. =

mv2
⇒ v = √

2K.E.

= √

2×9.4×10−20

9.1×10−31

= 4.54 × 105ms−1

Go to Question:

Prev Question

Next Question