If the radius of first Bohr orbit is $a_{0},$ then the radius of the third orbit would be

3 x a_{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

6 x a_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9 x a_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 / 9 x a_{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $3 x a_{0}$
$r_{n} = a_{0} \times n^{2}$ i.e., $r_{3} = a_{0} \times 3^{2} = 9 a_{0}$

Suggest Corrections

Similar questions

a_{0}

n^{t h}

2^{n d}

r_{2}

0.59 o A

H e^{+}

a_{0}

n^{th}

Join BYJU'S Learning Program