Question

Question 2.6 Find energy of each of the photons which

(i) correspond to light of frequency 3× 1015 Hz.

(ii) have wavelength of 0.50 Å.

Class XI Structure of Atom Page 65

Answer 1

according formula,
E = hv
where E is energy of photons
h is Plank's constant
v is frequency of photons

here,
(i)v = 3 × 10^15 Hz
h = 6.626 × 10^-34 Js
so, E = hv
= 6.626 × 10^-34 × 3 × 10^15
= 19.878 × 10^-19 J
= 1.9878 × 10^-18 J

we know,
E = hc/λ [ ∵ λ = cν ]
where, λ is wavelength of photons
c is the speed of light

here,
c = 3 × 10^8 m/s
λ = 0.5 A° = 5 × 10^-9 m

E = 6.626 × 10^-34 × 3 × 10^-8/5 × 10^-9
= 39.756 × 10^-16 J
= 3.9756 × 10^-15 J

Answer 2

Answer:

(ii) The answer is 3.98*10^-15.

Explanation:

By The formula hc/lambda

6.626*10^-34*3*10^8/0.50*10^-10

=3.98*10^-15