NEET Practice Questions (MCQs) with Answers & Solutions

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,

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

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

A particle of mass $M_1$ and charge $z_1$ possess initial velocity u, when it is at a large distance from the nucleus of an atom having atomic number $z_2$. At the distance of closest approach, the kinetic energy of particle is completely converted to potential energy. Mathematically, 

            $\frac{1}{2}{M}_1{u}^2=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{z_1{z}_2}{r_0}$

So the energy of the particle is directly proportional to $z_1{z}_2$.

Number of spectral lines obtained due to transition of electron from $n^{th}$ orbit to lower orbit is 

N = $\frac{n\left(n-1\right)}{2}$ and for maximum wavelength the difference between the orbits of the series should be minimum.

Number of spectral lines N = $\frac{n\left(n-1\right)}{2}$

$\Rightarrow \frac{n\left(n-1\right)}{2}=6$

or           $n^2-n-12=0$

or          $\left(n-4\right)\left(n+3\right)=0$

or                   n=4

Now as the first line of the series has the maximum wavelength, therefore electron jumps from the $4^{th}$ orbit to the third orbit.

Excitation energy is defined as the energy required to take the electron from ground level orbit to any higher order orbit (i.e, n= 2, 3, 4...)

Given, ground state energy of hydrogen atom 

                  $E_1=-13.6 eV$

Energy of electron in first excited state (i.e, n=2)

              $E_2=-\frac{13.6}{{\left(2\right)}^2}eV$

Therefore, excitation energy

       $∆E=E_2-E_1$

             =-$\frac{13.6}{4}-\left(-13.6\right)$

            =-3.4+13.6=10.2eV

The discharge of electricity through rarefied gases is an interesting phenomenon which can be systematically studies with the help of a discharge tube. In discharge tube collisions between the charged particles emitted from the cathode and the atoms of the gas results to the coloured glow in the tube .

(b) In an electric field, a force opposite to the direction of electric field acts on negatively charged particles (i.e. from lower potential to higher potential).

(d) 

Millikan

(c) $\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6\times {10}^{-19}}{9.1\times {10}^{-31}}=1.76\times {10}^{11}$C/kg