JEE Main 2019 — Physics solution

JEE Main2019Physics

A He⁺ion is in its first excited state. Its ionization energy is :

Solution

$E_{n} : \frac{- E _{0} z ^{2}}{n ^{2}} = \frac{- E _{0} \times 4}{4} = - E_{0}$

To ionise it E₀ energy must be supplied.

$∴ E_{0} = 13.6 e V$

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