Question

find would be the
approximate quantum number,n,for the circular orbit of hydrogen ,1×10*-5 cm in diameter?​

Answer 1

Answer:

ANSWER

Bhor radius is given as,

r

n

=0.529n

2

×10

−10

m

given that,diameter =10

−7

m

Thereforee, radius =10

−7

/2nm

Using bhor radius equation:

n

2

=10

−7

/0.529×10

−10

=1890.3

n=43.47 which

Explanation:

hope its helpful to you if this answer helps you so pls mark me as brainliest and follow me

Answer 2

Answer:

since...

$d = 1 \times {10}^{ - 5} cm$

$r = 1 \times ({10}^{ - 5} ) \div 2 \: cm$

$\: = 0.5 \times {10}^{ - 5} cm$

$= 5 \times {10}^{ - 6} cm$

$= 5 \times {10}^{ - 8} m$

$= 5 \times {10}^{ - 8} \times {10}^{10} angstrom$

Explanation:

Please check out the attachment.. For the rest!!

Hope it's helpful for you

Have a great day ahead