Question

The circumference (in m) of 3rd Bohr orbit in H-atom is 

(a) 3.0 x 10^-7
(b) 3.0 x 10^-8
(c) 3.0 x 10^-6 
(d) 3.0 x 10^-9

Answer 1

r = 0.529 * 10^-10 * n²/z

n is 3
z is 1

4.5 å
now multiply by 2 π

you get 2.7 nanometer
last option.

Answer 2

Answer:

The circumference (in m) of the 3rd Bohr orbit in H-atom is 3.0 x 10⁻⁹.

Explanation:

The circumference is calculated as,

C=2πr         (1)

Where,

C=circumferance of the orbit

r=radius of the orbit

And the radius of Bohr's orbit is calculated as,

r=0.529×n²A°    (2)

n=orbit number of Bohr's orbit

For the 3rd Bohr orbit of the hydrogen atom the radius is,

$\mathrm{r=0.529\times 3^2\times A^{\circ}}$

$\mathrm{r=4.761A^{\circ}}$    

$\mathrm{r=4.761\times 10^{-10}}$     (3)

Now the circumference of the 3rd Bohr orbit is calculated as,

$\mathrm{C=2\times \pi \times 4.761\times 10^{-10}}$               (π=3.14)

$\mathrm{C\approx 3\times 10^{-9}\ m}$        (4)

The circumference (in m) of the 3rd Bohr orbit in H-atom is 3.0 x 10⁻⁹.

Hence, the correct answer among the given options is (d).

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