Question

The energies E1 and E2 of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths ie λ1 and λ2 will be

Answer 1

Electromagnetic Radiation can be thought of as being made of packets of energy called Quanta, according to Planck's Theory.



The energy of each such "Quantum" of light is given by:

$E = h f$
where

$E$ is Energy
$h$ is Plack's Constant
$f$ is frequency of the Electromagnetic Radiation.


Now, frequency and wavelength are related as follows:

$c=\lambda \, f$

Here, $c$ is the speed of electromagnetic radiation. 

We have:

$c = \lambda \, f \\ \\ \\ \implies f = \frac{c}{\lambda}$

We can rewrite our Energy formula as:

$E = hf \\ \\ \\ \implies \boxed{E = \frac{hc}{\lambda}}$

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Coming to question. We have the data:

$E_1 = 25 \, \, eV \\ \\ E_2 = 50 \, \, eV$

We have to find the relation between corresponding wavelengths. 

We know:

$E = \frac{hc}{\lambda}$

So, we can write:
$E_1 = \frac{hc}{\lambda_1} \\ \\ \\ E_2 = \frac{hc}{\lambda_2}$

Dividing the first equation by the second one, we have:

$\frac{E_1}{E_2} = \frac{\frac{hc}{\lambda_1}}{\frac{hc}{\lambda_2}} \\ \\ \\ \implies \frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} \\ \\ \\ \implies \frac{25}{50} = \frac{\lambda_2}{\lambda_1} \\ \\ \\ \implies \frac{1}{2} = \frac{\lambda_2}{\lambda_1} \\ \\ \\ \implies \boxed{\lambda_1=2\lambda_2}$


Thus, we have our relation. The answer is Option (b)  $\lambda_1=2\lambda_2$


Hope it helps
Purva
Brainly Community

Answer 2

The answer is provided in the attachment.

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