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An electron in hydrogen atom jumps from third exited state to ground state how would the de broglie wavelength associated with electron change? Justify your answer.?
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Introduction

When an electron in a hydrogen atom jumps from a higher energy level to a lower energy level, it emits energy in the form of a photon. This emitted photon has a specific wavelength associated with it. According to de Broglie's hypothesis, every particle, including electrons, behaves like a wave and has a wavelength associated with it. The de Broglie wavelength of an electron can be calculated using the equation λ = h / p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the electron.

Explanation

1. Initial wavelength of the electron in the third excited state

Before the electron jumps to the ground state, let's calculate its initial de Broglie wavelength when it is in the third excited state. The energy levels in the hydrogen atom are given by the equation E = -13.6/n^2 eV, where E is the energy and n is the principal quantum number.

The energy of the electron in the third excited state (n = 3) can be calculated as follows:
E3 = -13.6/3^2 = -13.6/9 eV

The momentum of the electron can be calculated using the equation p = √(2mE), where p is the momentum and m is the mass of the electron. Substituting the values, we get:
p3 = √(2 * 9.1 x 10^-31 kg * (-13.6/9) * 1.6 x 10^-19 J/eV) = 1.52 x 10^-23 kg·m/s

Now, using the de Broglie wavelength equation, we can calculate the initial wavelength:
λ3 = h / p3 = (6.63 x 10^-34 J·s) / (1.52 x 10^-23 kg·m/s) ≈ 4.36 x 10^-11 m

2. Final wavelength of the electron in the ground state

When the electron jumps to the ground state (n = 1), its energy changes to:
E1 = -13.6/1^2 = -13.6 eV

The momentum of the electron in the ground state can be calculated as:
p1 = √(2 * 9.1 x 10^-31 kg * (-13.6) * 1.6 x 10^-19 J/eV) = 9.13 x 10^-23 kg·m/s

Using the de Broglie wavelength equation, we can calculate the final wavelength:
λ1 = h / p1 = (6.63 x 10^-34 J·s) / (9.13 x 10^-23 kg·m/s) ≈ 7.25 x 10^-10 m

3. Comparison of initial and final wavelengths

From the calculations, we can observe that the initial wavelength (λ3) is significantly smaller than the final wavelength (λ1). As the electron transitions from a higher energy level (third excited state) to a lower energy level (ground state), energy is released in the form of a photon. This emitted photon carries energy proportional to the energy difference between the two states. The energy of a photon is inversely proportional to its wavelength according to the
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An electron in hydrogen atom jumps from third exited state to ground state how would the de broglie wavelength associated with electron change? Justify your answer.?
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