Chemistry MCQs for NEET — Practice Questions with Answers

It represents

1.

It is found that wave functions ψ can be expressed as the product of two functions, one of which the radial part R(r) depends only on the distance from the nucleus, the other being the angular part f (θ, Φ ) depends only on the angles θ and Φ.

Probability of finding ${\mathrm{e}}^{-}$ is zero implies that ${\mathrm{\psi }}^2=0 \mathrm{or} \mathrm{\psi }=0$

$$\begin{gathered} \Rightarrow \left(\mathrm{\sigma }-1\right)=0 \Rightarrow \mathrm{\sigma }=1 \\ \mathrm{or} {\mathrm{r}}_1=\frac{{\mathrm{a}}_0}{2\mathrm{z}} \mathrm{or} \left({\mathrm{\sigma }}^2-8\mathrm{\sigma }+12\right)=0 \\ \mathrm{and} \left(\mathrm{\sigma }-6\right) \left(\mathrm{\sigma }-2\right)=0 \\ \mathrm{\sigma }=6, \mathrm{r}=\frac{3{\mathrm{a}}_0}{\mathrm{z}} \\ \mathrm{and} \mathrm{\sigma }=2, \mathrm{r}=\frac{{\mathrm{a}}_0}{\mathrm{z}} \\ {\mathrm{r}}_2=\frac{3{\mathrm{a}}_0}{\mathrm{z}} \end{gathered}$$

isodiaphers having  same  NEUTRON - PROTON

2-2 = 0.

8-8 = 0

3.

X rays is neutral, so cannot be deflected by electric and magnetic field

According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge +e. For an electron moving with a uniform speed in a circular orbit is a given radius, the centripetal force is provided by Columb force of attraction between the electron and the nucleus. The gravitational attraction may be neglected as the mass of electron and proton is very small.

so,

$$\begin{gathered} \frac{m{v}^2}{r}=\frac{k{e}^2}{r^2} \\ or m{v}^2=\frac{k{e}^2}{r}.......\left(1\right) \end{gathered}$$

where m=mass of electron

r=radius of electronic orbit

v=velocity of electron

Again,

$$\begin{gathered} mvr=\frac{nh}{2\mathrm{\pi }} \\ or v=\frac{nh}{2\mathrm{\pi mr}} \\ From eq. \left(1\right) we get \\ m{\left(\frac{nh}{2\mathrm{\pi mr}}\right)}^2=\frac{k{e}^2}{r} \\ r=\frac{n^2{h}^2}{4{\mathrm{\pi }}^2{\mathrm{kme}}^2}........\left(2\right) \\ U\sin g eq. \left(2\right) we get \\ {E}_K=\frac{k{e}^2}{2}\frac{4{\mathrm{\pi }}^2{\mathrm{kme}}^2}{n^2{h}^2} \\ =\frac{2{\mathrm{\pi }}^2{\mathrm{k}}^2{\mathrm{me}}^4}{n^2{h}^2} \\ \left(ii\right) Potential energy \\ {E}_P=-\frac{k\left(e\right)\times \left(e\right)}{r}=-\frac{k{e}^2}{r} \\ U\sin g eq. \left(2\right) we get \\ {E}_P=-k{e}^2\times \frac{4{\mathrm{\pi }}^2{\mathrm{kme}}^2}{n^2{h}^2} \\ =-\frac{4{\mathrm{\pi }}^2{\mathrm{k}}^2{\mathrm{me}}^2}{n^2{h}^2} \end{gathered}$$

Hence, total energy of the electron in the $n^{th}$ orbit

$$\begin{gathered} E=E_P+E_K=-\frac{4{\mathrm{\pi }}^2{\mathrm{k}}^2{\mathrm{me}}^4}{n^2{h}^2}+\frac{2{\mathrm{\pi }}^2{\mathrm{k}}^2{\mathrm{me}}^4}{n^2{h}^2}=-\frac{2{\mathrm{\pi }}^2{\mathrm{k}}^2{\mathrm{me}}^4}{n^2{h}^2} \\ =-\frac{13.6}{n^2}eV \end{gathered}$$

As we know that any orbit will be permitted only if it follow the following condition
mvr = n$\frac{h}{2\pi }$    (​mvr is the orbital angular momentum of Bohr orbit)
Rearranging the equation,
2$\pi$r = n$\frac{h}{mv}$ 
2$\pi$r = n $\lambda$  ( since $\frac{h}{mv}$ = $\lambda$ )

Therefore, the answer is n. Option 1

line spectra  which is  finger print of atom, is part of atomic spectra.

spectra can be of absorption or emission.