Question

The radius of Bohr's first orbit in H atom is 0.053nm. The radius of second orbit in He+ would be
A.0.0265nm
B.0.053nm
C.0.106nm
D.0.212nm

Answer 1

Answer:

r

n

=r

1

×

Z

n

2

r

2

=r

1

×

2

2

2

=0.106nm

Answer 2

Answer:

Concept:

The element hydrogen is made up of one proton and one electron, and it often takes the form of a gas. The most prevalent element in the universe, hydrogen, which makes up around 75% of all ordinary stuff, was produced during the Big Bang. A two protons and two neutrons, surrounded by two electrons, nucleus makes up the element helium, which is often found as a gas. A quarter of all atoms in the universe are made of helium, which is the second most prevalent element in the cosmos after hydrogen. The majority of the helium in the cosmos was produced during the Big Bang, but it was also a byproduct of the fusion of hydrogen in stars.

Explanation:

Given :

Hydrogen atom = 0.053 A°

To Find :

The radius of second orbit in He+

Solution:

By atom number it says,

Step 1 :

Single electron

$r_{n} = r_{0}$ × $\frac{n^{2} }{z}$ A°

$r_{0} =$   0.053 A°

Step 2 :

- He+  = 2 - 1 = 1$e^{-}$

$r_{z}$ = 0.53 × $\frac{(2^{2} )}{z}$  A°

$r_{z}$    =  2 ×  0.053 A°

$r_{z}$   = 1.06 A°

Step 3 :

1nm = $10^{-9}$ nm

$r_{z}$ = 1.06 A° × $10^{-10}$ m

$r_{z}$    = 0.106 × $10^{-9}$m

$r_{z}$   = 0.106 nm.

The radius of second orbit in He+ would be is 0.106 nm.

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