JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Medium

1 of 1

Application +4 | -1

The radius of the second Bohr orbit for hydrogen atom is (Planck’s constant

\left( h \right) = 6.6262 \times {10^{ - 31}}Js;

mass of electron

= 9.1091 \times {10^{ - 31}}\,kg;

charge of electron

\left( e \right) = 1.60210 \times {10^{ - 19}}\,C;

permittivity of vacuum

\left( {{ \in _0}} \right) = 8.854185 \times {10^{ - 12}}k{g^{ - 1}}{m^{ - 3}}{A^2}

)

View answer and explanation

Correct option: D

2.12\,\mathop \mathrm{A}\limits^ \circ

Option D (

2.12\,\mathop \mathrm{A}\limits^ \circ

) is correct because it is calculated using the formula for the radius of the nth Bohr orbit for hydrogen atom.

More Questions from STRUCTURE OF ATOM

Practice All Chapter MCQs →

n^{\text {th }}

\mathrm{A}

\times 10^{12}

\frac{1}{{ }^{\prime} \mathrm{B}^{\prime}}

\ell = 1,\,\,m = 0

1.25 \times {10^{ - 13}} \times {A^{1/3}}\mathrm{cm}.