Question

The ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is(a) 2 : 1(b) 4 : 1(c) 8 : 1(d) 16 : 1

Answer 1

Answer:The ratio of the area is 16:1.

Explanation:

We know that,

The square of radius of the orbit is directly proportional to power of four of the  number of state.

$r^2\propto n^4$

Where, n is the number of state

The area of the hydrogen for excited state

$A_{1} \propto n^4_{1}$

Here, n = 2 for excited state

The area of the hydrogen for ground state

$A_{0} \propto n^4_{0}$

Here, n = 1 for ground state

Now, the ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is

$\dfrac{A_{1}}{A_{0}}= (\dfrac{n_{1}}{n_{0}})^4$

$\dfrac{A_{1}}{A_{0}}= (\dfrac{2}{1})^4$

$\dfrac{A_{1}}{A_{0}}= \dfrac{16}{1}$

Hence, The ratio of the area is 16:1.

Answer 2

Answer:

D) 16 : 1

Explanation:

The radius of hydrogen atom is directly proportional to the square of its principle quantum number,

Thus, for the first excited state will be = n= 2

and the ground state will be n = 1

Therefore,

R= 4 as (n²= 2²)

R'=1 (as n = 1² )

Thus the ratio used will be -

= πr²/ πR'²

= 16/1

Therefore, the ratio of areas between the electron orbits for the first excited state to the ground state for the hydrogen atom is 16 : 1