In Hydrogen atom , if an electron jump from n2 level into n1 level . Prove that the wave number (reciprocal of wave length) of emitted radiation is :
1/λ=me4 /8ε02h3c (1/n22 -1/n12)
To prove this equation, we need to make use of the Bohr model of the hydrogen atom and the relationship between the energy levels and the wavelengths of emitted light. Let's break it down step by step:
Step 1: Start with the Bohr model equation for the energy levels of an electron in a hydrogen atom
E = -me^4 / (2*ε₀^2 *h^2) * (1/n^2)
Where:
- E represents the energy level of the electron
- me represents the mass of the electron
- e represents the elementary charge
- ε₀ represents the vacuum permittivity
- h represents the Planck's constant
- n represents the principal quantum number (energy level)
Step 2: Calculate the energy difference between the initial (n2) and final (n1) energy levels
ΔE = E(n2) - E(n1)
= -me^4 / (2ε₀^2*h^2) * (1/n2^2) + me^4 / (2ε₀^2*h^2) * (1/n1^2)
= (me^4 / 2ε₀^2*h^2) * (1/n1^2 - 1/n2^2)
Step 3: Apply the relationship between energy and wavelength for electromagnetic radiation
ΔE = hc/λ
Where:
- ΔE represents the energy difference between the initial and final energy levels
- c represents the speed of light
- λ represents the wavelength
Step 4: Substitute hc/λ for ΔE in the previous equation
hc/λ = (me^4 / 2ε₀^2*h^2) * (1/n1^2 - 1/n2^2)
Step 5: Rearrange the equation to solve for the reciprocal of the wavelength (1/λ)
1/λ = (me^4 / 2ε₀^2*h^2) * (1/n1^2 - 1/n2^2) / hc
Step 6: Simplify the equation further by rearranging terms and cancelling constants
1/λ = me^4 / 8ε₀^2 * h^3 * c * (1/n1^2 - 1/n2^2)
And there you have it! The final equation proves that the wave number (reciprocal of wavelength) of the emitted radiation from the hydrogen atom is given by 1/λ = me^4 / 8ε₀^2 * h^3 * c * (1/n1^2 - 1/n2^2).