Question

The radius of the nth orbit of a single electron species is $0.132\times {\mathrm{n}}^2\mathrm{angstrom}$. Identify the element.

Answer 1

${\mathrm{r}}_{\mathrm{n}}=\frac{{\mathrm{n}}^2{\mathrm{h}}^2}{4{\mathrm{\pi }}^2{\mathrm{Zme}}^2}$is the equation to calculate the radius of the nth orbit of the single electron species.

where h= Planck's constant(6.625x10^-34Js),

n= number of orbits

m= mass of an electron(9.1 x 10^-28g)

e= charge of electron(1.6x 10^-19C)

Z= atomic number of elements.

By placing all these values, an Equation to calculate the radius of the nth orbit of the hydrogen atom$\left({\mathrm{r}}_{\mathrm{n}}\right)=0.529\times \frac{{\mathrm{n}}^2}{\mathrm{Z}}\mathrm{angstrom}$.

If the radius of the nth orbit of a single electron species is$0.132\times {\mathrm{n}}^2\mathrm{angstrom}$ then $\left({\mathrm{r}}_{\mathrm{n}}\right)=0.529\times \frac{{\mathrm{n}}^2}{\mathrm{Z}}\mathrm{angstrom}=0.132\times {\mathrm{n}}^2\mathrm{angstrom}$

The atomic number of the element(Z)=4

The element must be Beryllium.