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Question

An electron in a hydrogen like atom makes a transition from a state in which its de-Broglie wavelength is $\lambda_1$ to a state in which its de-Broglie wavelength is $\lambda_2$, then wavelength of photon generated due to this transition will be ___________.

The solution for this problem as given in my book is as follows:

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In the first step, it is given that the energy difference is just equal to the kinetic energy difference. How is this possible? Why have they neglected the potential energy in calculating the energy difference since the two stationary orbits in Bohr's model have different potential as well as kinetic energies.

Please clarify my doubt.

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closed as off-topic by AccidentalFourierTransform, John Rennie, stafusa, glS, Aaron Stevens Sep 3 at 14:30

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In the Bohr model of the hydrogen atom, the total (kinetic + potential) energy turns out to be the negative of the kinetic energy. For example, in the ground state, the kinetic energy is +13.6 eV, the potential energy is -27.2 eV, and the total energy is -13.6 eV.

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