Question

A hydrogen- like atom has ground state binding energy $122.4eV$. Then

• A. its atomic number is $3$
• B. an electron with $90eV$ can excite it to a higher state
• C. an $80eV$ electron can excite it to a higher state
• D. an electron of $8.2eV$ and a photon of $91.8eV$ emitted when a $100eV$ electron interacts with it

Answer 1

Answer: (A), (D)

The correct options are
A its atomic number is $3$
D an electron of $8.2eV$ and a photon of $91.8eV$ emitted when a $100eV$ electron interacts with it

$E_n=\frac{-z^{2}13.6eV}{n^2}$

for ground state, $n=1$

$-122.4=-Z^{2}13.6eV$

When a proton interact that

$E_P=hV=E_1-E_0=Z^{2}13.6eV[\frac{1}{1^2}-\frac{1}{2^2}]$

$=9\times 13.6eV\left[3/4\right]$

$=91.8eV$

So the Alomic number is $3$ , an photon of $91.8eV$ and electron of $8.2eV$ are emitted when 100eV electron interacts $e^{-}$ interct with this atom.