Radius of the second Bohr orbit of a singly ionised helium atom is

  • Option 1)

    0.53 A \degree

  • Option 2)

    1.06 A \degree

  • Option 3)

    0.265 A \degree

  • Option 4)

    0.132 A \degree

 

Answers (1)

As we learned in concept

Radius of nth orbital -

r_{n}= \frac{\epsilon _{0}n^{2}h^{2}}{\pi mZe^{2}}

 

- wherein

r_{n}\alpha \: \frac{n^{2}}{Z}

\frac{\epsilon_{0}h^{2}}{\pi me^{2}}= 0.529A^{\circ}

 

 R = R_{o}\frac{n^{2}}{Z}

n=2 and Z=2

\therefore R-R_{o}\:.\:\frac{4}{2}=2R_{o}=1.06A^{\circ} (Since R_{O}=0.53A^{\circ})


Option 1)

0.53 A \degree

Option 2)

1.06 A \degree

Option 3)

0.265 A \degree

Option 4)

0.132 A \degree

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