Collision of Photon with Helium Ion
When a photon of energy 70 electron-volt collides with a helium ion (He+), the ion can get ionized. This process can result in the emission of an electron with a certain kinetic energy.

Maximum Kinetic Energy of Emitted Electron
The maximum kinetic energy of the emitted electron can be calculated using the energy conservation principle. The energy of the incident photon is used to ionize the helium ion and provide kinetic energy to the emitted electron.
The maximum kinetic energy of the emitted electron can be calculated using the equation:
\[K_{max} = E_{photon} - E_{ionization}\]
Where:
- \(K_{max}\) is the maximum kinetic energy of the emitted electron
- \(E_{photon}\) is the energy of the incident photon (70 electron-volt)
- \(E_{ionization}\) is the ionization energy of the helium ion

Explanation
When the photon collides with the helium ion, it transfers its energy to the ion. This energy is used to remove an electron from the helium ion, resulting in ionization. The remaining energy is then transferred to the emitted electron in the form of kinetic energy.
By calculating the difference between the energy of the incident photon and the ionization energy of the helium ion, we can determine the maximum kinetic energy of the emitted electron.
In this scenario, the maximum kinetic energy of the emitted electron can be calculated by subtracting the ionization energy of the helium ion from the energy of the incident photon (70 electron-volt).

From Bohr's theory for a hydrogen like atom

En =−Z^2/n^2×13.6 eV
For He+,
Z=2 and n=1
⇒E1=−54.4 eV

By conserving energy

K.E of electron = Energy of Photon - E1
=70−54.4
=15.6 eV