NEET Practice Questions (MCQs) with Answers & Solutions

(c) Key idea Excess energy of $e^{-}$ appears as photon.

 From Rydberg's formula,

           $$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{{{\displaystyle n}}_1^2}-\frac{1}{n_1^2}\right)=R\left(\frac{1}{2^2}-\frac{1}{2^2}\right)=\frac{{\displaystyle 5R}}{{\displaystyle 36}} \\ \frac{{\displaystyle 1}}{{\displaystyle \lambda }}=R \left(\frac{{\displaystyle 1}}{{\displaystyle 3^2}}-\frac{{\displaystyle 1}}{{\displaystyle 4^2}}\right)=\frac{{\displaystyle 7R}}{{\displaystyle 144}} \\ \frac{{\displaystyle 1}}{{\displaystyle \lambda }}/\frac{{\displaystyle 1}}{{\displaystyle \lambda \text{'}}}=\frac{{\displaystyle 5R}}{{\displaystyle 36}}+\frac{{\displaystyle 7R}}{{\displaystyle 144}} \\ \Rightarrow \frac{{\displaystyle \lambda \text{'}}}{{\displaystyle \lambda }}=\frac{{\displaystyle 5R}}{{\displaystyle 36}}\times \frac{{\displaystyle 144}}{{\displaystyle 7R}}=\frac{20}{7} \\ \Rightarrow \lambda \text{'}=\frac{20}{7}\lambda \end{gathered}$$

(b) Given, Rydberg constant, $R={10}^7{m}^{-1}$

$\because$ For last time in Balmer series, $n_2=\infty ,n_1=2.$

As we know that

    $$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{{\displaystyle 1}}{{\displaystyle n_2^2}}\right)\Rightarrow \frac{{\displaystyle 1}}{{\displaystyle \lambda }}={10}^7\left(\frac{{\displaystyle 1}}{{\displaystyle 2^2}}-\frac{{\displaystyle 1}}{{\displaystyle \infty }}\right) \\ \Rightarrow \overline{V}=\frac{{\displaystyle 1}}{{\displaystyle \lambda }}=\frac{{10}^7}{4}=0.25\times {10}^7{m}^{-1} \end{gathered}$$

(d) When an $\alpha$-particle of mass m moving with velocity v bombards on a heavy nucleus of charge Ze, then there will be no loss of energy as in case, initial kinetic enorgy of $\alpha$-particle potential energy of α-particle at closest approach

=> $\frac{1}{2}$mv²=$\frac{2z{e}^2}{4\mathrm{\pi }{\epsilon }_0{r}_0}$

r_o∝$\frac{1}{m}$

This is the required closest approach to $\alpha$-particle from the nucleus

In hydrogen atom, wavelength of characteristic spectrum

$\frac{1}{\lambda }$=Rz² [1/n₁²-1/n₂²]

For Lyman series n₁=1 , n₂=2

$\frac{{\displaystyle 1}}{{\displaystyle {\lambda }_1}}= =R{z}^2[1/{\left(1\right)}^2-1/{\left(2\right)}^2]$…(i)

For Balmer series n₁=2, n₂=3

$\frac{{\displaystyle 1}}{{\displaystyle {\lambda }_2}}==R{z}^2 [1/{\left(2\right)}^2-1{\left(3\right)}^2]$… (ii)

Dividing Eq. (ii) by Eq. (i) we get

$\frac{{\displaystyle {\lambda }_1}}{{\displaystyle {\lambda }_2}}$=5/36 x 4/3 =5/27

(c)

$$\begin{gathered} Energy provided to the ground state electron: \\ E=\frac{hc}{\lambda }=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}}=12.75eV \\ It means the electron jumps to 3rd excited state ( n=4) \\ No. of spectral lines=\frac{n\left(n-1\right)}{2}=6 \end{gathered}$$

(a) Wavelength for Lyman series 

λ_L=1/R(1-1/4)=4/3R

and wavelength for Balmer series

λ_B=1/R(1/4-1/9)=1/R(5/36)=36/5R

   ∴λ_L/λ_B=4/3R x 5R/36=5/27

       =>λ_L:λ_B=5:27  

Here, for wavelength ${\lambda }_1$

$n_1=4 and n_2=3$

and for ${\lambda }_2, n_1=3 and n_2=2$

We have $\frac{hc}{\lambda }=-13.6\left[\frac{1}{n_2^2}-\frac{{\displaystyle 1}}{{\displaystyle n_1^2}}\right]$

So, for ${\lambda }_1$

$\Rightarrow \frac{hc}{{\lambda }_1}=-13.6\left[\frac{1}{{\left(4\right)}^2}-\frac{1}{{\left(3\right)}^2}\right]$

         $\frac{hc}{{\lambda }_1}=13.6\left[\frac{7}{144}\right] ...\left(i\right)$

Similarly, for ${\lambda }_2$

$\Rightarrow \frac{hc}{{\lambda }_2}=-13.6\left[\frac{1}{{\left(3\right)}^2}-\frac{1}{{\left(2\right)}^2}\right]$

         $\frac{hc}{{\lambda }_2}=13.6\left[\frac{5}{36}\right] ...\left(ii\right)$

Hence, from Eqs. (i) and (ii), we get 

                     $\frac{{\lambda }_1}{{\lambda }_2}=\frac{20}{7}$

$$\begin{gathered} \frac{1}{\lambda }=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right) \\ Here n_1=1 and n_2=5 \\ Hence \frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right) =\frac{24}{25}R \\ Now photon energy- \\ E=\frac{hc}{\lambda }=\frac{24}{25}hcR \\ Here the momentum of photon=momentum of atom \\ Thus P=\frac{E}{c}=\frac{24}{25}hR \\ Velocity of atom v=\frac{P}{m}=\frac{24hR}{25m} \end{gathered}$$

Energy released from emission of electron

           $E=\left(-3.4\right)-\left(-13.6\right)$

              $=10.2 eV$

From photo electric equation.

Work function

   $\varphi =E-eV=hv$

   $v=\frac{E-eV}{h}$

     =$\frac{\left(10.2-3.57\right)e}{6.67\times {10}^{-34}}$

v=$\frac{6.63\times 1.6\times {10}^{-19}}{6.67\times {10}^{-34}}$

$=1.6\times {10}^{15}Hz$

Infrared radiation is found in Paschan, Brackett

and pfund series and it is obtain when electron

transition occur from high energy level to 

minimum third level.