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NEET CHEMISTRYEasy

The energies E₁ and E₂ of two radiations are 25 eV and 50 eV respectively. The relation between their wavelengths i.e., \lambda ₁ and \lambda ₂ will be:

A

\lambda ₁ = 2\lambda ₂

B

\lambda ₁ = 4\lambda ₂

C

\lambda ₁ = ½\lambda ₂

D

\lambda ₁ = \lambda ₂

Step-by-Step Solution

The energy (EE) of a photon or radiation is inversely proportional to its wavelength (λ\lambda) according to the Planck-Einstein relation: E=hcλE = \frac{hc}{\lambda} where hh is Planck's constant and cc is the speed of light.

Since hchc is constant, we can write the ratio: E1E2=λ2λ1\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}

Given: E1=25 eVE_1 = 25 \text{ eV} E2=50 eVE_2 = 50 \text{ eV}

Calculation: 2550=λ2λ1\frac{25}{50} = \frac{\lambda_2}{\lambda_1} 12=λ2λ1\frac{1}{2} = \frac{\lambda_2}{\lambda_1} λ1=2λ2\lambda_1 = 2\lambda_2

Thus, the wavelength of the first radiation is twice that of the second.

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