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Question
There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength?
Solution
1. X-rays were discovered by scientist Roentgen which is why they are also called Roentgen rays.
2. Roentgen discovered that when the pressure inside a discharge tube is kept at 10“3 mm of Hg and the potential difference is kept at 25 kV, then some unknown radiations (X-rays) are emitted by the anode.
3. There are three essential requirements for the production of X-rays.
- A source of electron
- An arrangement to accelerate the electrons
- A target of suitable material of high atomic weight and high melting point on which these high-speed electrons strike.
Here in this problem, total energy will be constant.
Let us assume the wavelength of X-rays is λ1 and the wavelength of visible light is λ2.
Given, P = 100 W
λ1 = 1 nm
and λ2 = 500 nm
Let n1 and n2 be the number of photons of X-rays and visible light emitted from the two sources per sec.
So, `E/t = P = n_1 (hc)/λ_1 = n_2 (hc)/λ_2`
⇒ `n_1/λ_1 = n_2/λ_2`
⇒ `n_1/n_2 = λ_1/λ_2 = 1/500`
Important point: The wavelength of characteristics of X-ray doesn't depend on accelerating voltage. It depends on the atomic number (Z) of the target material. In characteristic X-ray spectrum λKα < λLα < λMα and Vkα < VLα < VMα. Also λKα < λKβ < λKγ
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