An electron in the hydrogen atom initially in the fourth excited state makes a transition to 
energy state by emitting a photon of energy 2.86 eV. The integer value of n will be _________.
Text Solution
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To find the integer value of 
for which an electron transitions from the fourth excited state in a hydrogen atom, thus emitting a photon with an energy of 2.86 eV , we can follow these steps:
We use the formula for the energy difference associated with electron transitions in a hydrogen atom:
However, in this question, it seems there's a typo in the original explanation, as both indices are given as 
. To correct it, we should use this formula:
where 
(the fifth energy level or fourth excited state) and 
(the state to which the electron transitions).
Given the photon's energy is 2.86 eV, set up the equation:
Solve for 
:
Calculate the value:
Find 
:
Consequently:
Thus, the electron transitions to the 
energy state.
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