Physics MCQs for NEET — Practice Questions with Answers

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.

Lyman series for H-ion

                    $\frac{\mathrm{hc}}{\mathrm{\lambda }}=\mathrm{Rhc}\left(\frac{1}{1^2}-\frac{1}{2^2}\right)$

and for H-like ion 

                     $\frac{\mathrm{hc}}{\mathrm{\lambda }}=Z^2\mathrm{Rhc}\left(\frac{1}{2^2}-\frac{1}{4^2}\right)$

$$\begin{gathered} \therefore \left(\frac{1}{1^2}-\frac{1}{2^2}\right)={\mathrm{Z}}^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ \left(1-\frac{1}{4}\right)={\mathrm{Z}}^2\left(\frac{1}{4}-\frac{1}{16}\right) \\ \mathrm{Z}=2 \end{gathered}$$

Energy E of an atom with principal quantum number n is given by E=$\frac{-13.6}{n^2}{Z}^2$ for first excited state n=2 and for ${\mathrm{He}}^{+} \mathrm{Z}=2$

$\Rightarrow E=\frac{-13.6\times {\left(2\right)}^2}{{\left(2\right)}^2}$

               =-13.6 eV

A $\alpha ‐$particle of mass m possesses initial velocity v, when it is at a large distance from the nucleus of an atom having atomic number Z. At the distance of closest approach, the kinetic energy of $\alpha ‐$particle is completely converted into potential energy. Mathematically,

            $\frac{1}{2}m{v}^2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(2e\right){\displaystyle \left(Ze\right)}}{r_0}$

           $r_0=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{2Z{e}^2}{{\displaystyle \frac{1}{2}}m{v}^2}$