Question

energy difference between 5th and 4th orbit is a.0.66eV b.1.9eV c.0.30eV d.10.2 eV
​

Answer 1

Answer:

The energy difference between the 5th and 4th orbit is 0.306 eV i.e.option(c).

Explanation:

The energy difference is calculated as,

$\mathrm{\Delta E=13.6(\frac{1}{n_1^2}- \frac{1}{n_2^2}) }$               (1)

Where,

ΔE=energy difference between the two levels

n₁=lower energy state

n₂=higher energy state

From the question, we can interpret that

The lower energy state(n₁)=4

Higher energy state(n₂)=5

By inserting the numbers in equation (1), we receive;

$\mathrm{\Delta E=13.6(\frac{1}{4^2}- \frac{1}{5^2}) }$

$\mathrm{\Delta E=13.6(\frac{1}{16}- \frac{1}{25}) }$

$\mathrm{\Delta E=13.6(\frac{25-16}{16\times 25}) }$

$\mathrm{\Delta E=0.306\ eV }$

So, the energy difference between the 5th and 4th orbit is 0.306 eV i.e.option(c).

#SPJ2

Answer 2

The energy difference between the 5th and 4th orbit is 0.306 eV

correct answer is Option(c).

Explanation:

Range of energy levels is a quantized energy spectrum of a system.

When we are calculating the rotational energy levels formula the Bohr model of the hydrogen atom is  considered.

The energy difference is calculated as,

Δ$E=13.6 (\frac{1}{n1^{2} } -\frac{1}{n2^{2} } )$

Where,

ΔE=energy difference between the two levels

n₁=lower energy state

n₂=higher energy state

From the question, we can interpret that

The lower energy state(n₁)=4

Higher energy state(n₂)=5

By inserting the numbers in equation (1), we receive;

Δ$E=13.6 (\frac{1}{4^{2} } -\frac{1}{5^{2} } )$

Δ$E=13.6 (\frac{1}{16} -\frac{1}{25 } )$

Δ$E= 0.306 eV$

Energy difference between 5th and 4th orbit is Δ$E= 0.306 eV$

#SPJ2