Atomic Structure

Metals emit electrons when light shines upon them. The phenomenon is called as photoelectric effect and the electrons emitted in such manner are called photoelectrons.

Relation between kinetic energy and frequency is kinetic energy $=\phi-h \mathrm{v}$

Here $\phi=$ metal work function

Kinetic energy of photoelectrons depends on frequency of incident light and metal work function. $\mathrm{Cu}$ has maximum metal work function's value.

Hence, $\mathrm{Cu}$ have minimum kinetic energy of ejected photoelectrons than alkali metals.

We know that, energy $=h c / \lambda$

Here, $h=$ Planck's constant (Js )

$c=$ speed of light $(\mathrm{m} / \mathrm{s})$

$\lambda=$ wavelength $(\mathrm{m})$

Now for energy $E_1=25 \mathrm{eV}$

$E_1=h c / \lambda_1 \Rightarrow 25=h c / \lambda_1$ ..... (i)

Similarly for energy,

$E_2=h c / \lambda_2$

$100=h c / \lambda_2$ ..... (ii)

By divide $E_1$ and $E_2$, we get

$\begin{aligned} & 25 / 100=\lambda_2 / \lambda_1 \Rightarrow 1 / 4=\lambda_2 / \lambda_1 \\ & \lambda_1: \lambda_2=4: 1 \text { or } \lambda_1=4 \lambda_2 \end{aligned}$

Let us consider, particle $A$ having velocity $=v_A$ and particle $B$ having velocity $=v_B$

According to de-Broglie wavelength, $\lambda=\frac{h}{m v}$

here, $h=$ Planck constant

$m=$ mass of electron

$v=$ velocity of particle

Now, $\lambda_A=\frac{h}{m_A v_A}$ and $\lambda_B=\frac{h}{m_B v_B}$

Wavelength of $A$ is found to be double than of $B$.

$\text { i.e. } \lambda_A=2 \times \lambda_B$ ..... (i)

Put value of $\lambda_A$ and $\lambda_B$ in Eq. (i)

$\begin{array}{rlrl} 2 \times \frac{h}{m_B v_B} & =\frac{h}{m_A v_A} & & \left(\text { For } v_A=v_B\right) \\ \frac{2}{m_B} & =\frac{1}{m_A} & {\left[m_A=\frac{m_B}{2}\right]}\end{array}$

Here, mass of $A$ is half that of $B$.

Number of electrons in lithium ion (Li⁺) is 2.

Number of electrons in helium is 2.

The spectrum of an atom depend on the number of electrons present in it.

Li$^+$ (Z = 3)

He (Z = 2)

(Both have same electrons = 2)

Bohr radius of hydrogen atom is first orbit in H-atom.

Bohr radius $({r_n}) = \left( {{{{n^2}} \over Z}} \right)\left( {{{{n^2}} \over {4{\pi ^2}m{e^2}}}} \right)$

${r_n} = {{{n^2}} \over Z} \times 0.529\,\mathop A\limits^o $

For first orbit $(n = 1),{r_1} = \left( {{1 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (i)

For second orbit $(n = 2),{r_2} = \left( {{4 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ ..... (ii)

For third orbit $(n = 3),{r_3} = \left( {{9 \over Z}} \right) \times 0.529\,\mathop A\limits^o $ .... (iii)

Now divided (iii) by (i), we get

${r_3} = 9{r_1}$

The radius of 3rd orbit is 9 times the radius of 1st orbit.