Q17 / 201 · NTA Abhyas (NEET) 2020 · Physics · Atomic Physics

Ionization energy of $H e^{+}$ ion at minimum energy position is

A
$13.6 e V$
B
$27.2 e V$
Correct answer
C
$54.4 e V$
D
$68.0 e V$

Solution

The energy of helium ions.

$E_{n} = - \frac{13 .6 Z^{2}}{n^{2}} eV$

In minimum position, $n = 1$ .

For ${He}^{+}, Z = 2$ .

$E = \frac{- 13 .6 {\times (2)}^{2}}{1} eV$

$E =- 54.4 eV$

Ionization Energy = $- E = 54.4 eV$

Hence, the correct answer is option: 'C'.

More Atomic Physics Questions

The ratio of the maximum wavelength of the Lyman series in hydrogen spectrum to the maximum wavelength in the Paschen series is
K α wavelength emitted by an atom of atomic number Z = 11 is λ . Find the atomic number for an atom that emits K α radiation with wavelength
The binding energy per nucleon of deuteron 1 2 H and helium nucleus 2 4 He is 1.1 Me V and 7 Me V respectively. If two deuteron nuclei react
We wish to observe an object which is 2.5 Å in size. The minimum energy photon that can be used is
The ionisation energy of 10 time ionised sodium atom is
The graph given below, shows the variation of f vs z for characteristics X- rays. Lines 1,2,3 and 4 shown in the graph corresponds to any on

Ionization energy of He+ ion at minimum energy position is

Solution

More Atomic Physics Questions

Ionization energy of $H e^{+}$ ion at minimum energy position is