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MHTCET 15 May 2024 Shift 2 Paper

Question : 60 of 150

Marks: +1, -0

\mathrm{He}^{+}

Solution:

Concept:

For a hydrogen-like ion, the radius of the

n^{\text{th}}

Bohr orbit is given by

r_n = \frac{52.9\,n^2}{Z}\ \mathrm{pm}

Explanation:

\mathrm{He}^{+}

is a hydrogen-like species with atomic number

Z = 2

For the first orbit,

n = 1

Substitute

n = 1

and

Z = 2

in the formula:

r_1 = \frac{52.9 \times (1)^2}{2}\ \mathrm{pm}

r_1 = \frac{52.9}{2} = 26.45\ \mathrm{pm}

Hence, the radius of the first orbit of

\mathrm{He}^{+}

26.45\ \mathrm{pm}

Answer:

26.45\ \mathrm{pm}

None of the given options matches this value exactly.

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