Radius of a certain orbit of hydrogen atom is 8.48 \mathop A\limits^o . If energy of electron in this orbit is E / x . then x= \underline\hspace2cm (Given a_0=0.529\mathop A\limits^o , E= energy of electron in ground state).

Atoms and Nuclei question

2024 · 6 Apr · Shift 1 · Q85

2024 · 6 Apr · Shift 1 · Q85

View written solutionFree

Correct answer: 16

For hydrogen atom, radius of the $n^{\text{th}}$ orbit is

r_n = n^2 a_0

Given:

r_n = 8.48\,\mathring{A}, \qquad a_0 = 0.529\,\mathring{A}

So,

n^2 = \frac{r_n}{a_0} = \frac{8.48}{0.529}

Now,

\frac{8.48}{0.529} \approx 16.03 \approx 16

Hence,

n^2 = 16 \Rightarrow n=4

Energy of electron in the $n^{\text{th}}$ orbit is

E_n = \frac{E_1}{n^2}

where $E_1$ is the ground state energy in magnitude form.

Given in the question that energy in this orbit is $E/x$ , where $E$ is the energy in ground state. So,

\frac{E}{x} = \frac{E}{n^2}

Therefore,

x = n^2 = 16

\boxed{x=16}

2024 · 6 Apr · Shift 1 · Q85

2024 · 6 Apr · Shift 1 · Q85

View written solutionFree

Correct answer: 16

For hydrogen atom, radius of the $n^{\text{th}}$ orbit is

r_n = n^2 a_0

Given:

r_n = 8.48\,\mathring{A}, \qquad a_0 = 0.529\,\mathring{A}

So,

n^2 = \frac{r_n}{a_0} = \frac{8.48}{0.529}

Now,

\frac{8.48}{0.529} \approx 16.03 \approx 16

Hence,

n^2 = 16 \Rightarrow n=4

Energy of electron in the $n^{\text{th}}$ orbit is

E_n = \frac{E_1}{n^2}

where $E_1$ is the ground state energy in magnitude form.

Given in the question that energy in this orbit is $E/x$ , where $E$ is the energy in ground state. So,

\frac{E}{x} = \frac{E}{n^2}

Therefore,

x = n^2 = 16

\boxed{x=16}