Chemistry MCQs for NEET — Practice Questions with Answers

4. 

Bohr's Theory is based on fixed energy difference of shells

In the photoelectric effect, the kinetic energy of the emitted photoelectrons is directly proportional to the frequency of the incident light, not its wavelength, velocity, or any property of the metal. This relation is given by Einstein's photoelectric equation.

$$\begin{gathered} hv=h{v}_0+KE \\ hv=h{v}_0+e{V}_0 \\ {V}_0=\frac{hv}{e}-\frac{h{v}_0}{e} \left(y=mx+c\right) \\ \Rightarrow m=\frac{h}{e} \end{gathered}$$

3.

The minimum frequency required for the emission of electrons is called threshold frequency below which no photoelectron is emitted. ... The minimum energy required for the emission of photoelectrons from a metal is called work function of that metal

As per Einstein's equation of photoelectric effect $\mathrm{hv}={\mathrm{hv}}_0+\mathrm{K}.\mathrm{E}.$

$$\begin{gathered} \therefore \frac{1}{2}{\mathrm{mv}}^2=\mathrm{hv}-{\mathrm{hv}}_0=\frac{\mathrm{hc}}{\mathrm{\lambda }}-\frac{\mathrm{hc}}{{\mathrm{\lambda }}_0} \\ {\mathrm{v}}^2=\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{1}{\mathrm{\lambda }}-\frac{1}{{\mathrm{\lambda }}_0}\right); \\ \mathrm{v}={\left[\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{1}{\mathrm{\lambda }}-\frac{1}{{\mathrm{\lambda }}_0}\right)\right]}^{1/2} \\ \Rightarrow {\left[\frac{2\mathrm{hc}}{\mathrm{m}}\left(\frac{{\mathrm{\lambda }}_0-\mathrm{\lambda }}{{\mathrm{\lambda \lambda }}_0}\right)\right]}^{1/2} \end{gathered}$$

$$\begin{gathered} \mathrm{hv}={\mathrm{hv}}_0+{\mathrm{eV}}_0; {\mathrm{eV}}_0=\mathrm{hv}-{\mathrm{hv}}_0 \mathrm{or} \\ {\mathrm{V}}_0=\frac{\mathrm{h}}{\mathrm{e}}\mathrm{v}-\frac{\mathrm{h}}{\mathrm{e}}{\mathrm{v}}_0; {\mathrm{slope}}_1=\frac{\mathrm{h}}{\mathrm{e}} \\ \mathrm{Similarly}, \mathrm{hv}={\mathrm{hv}}_0+{\mathrm{K}}_{\max } \\ \mathrm{or} {\mathrm{K}}_{\max }=\mathrm{hv}-{\mathrm{hv}}_0; \\ {\mathrm{slope}}_2=\mathrm{h}, \frac{{\mathrm{slope}}_2}{{\mathrm{slope}}_1}=\frac{\mathrm{h}}{\mathrm{h}/\mathrm{e}}=\mathrm{e} \end{gathered}$$

wavelength is inversely proportional to mass.

$$\begin{gathered} \because \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right); {\mathrm{n}}_1=1, {\mathrm{n}}_2=?; \\ \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{1}-\frac{1}{{\mathrm{n}}_2^2}\right) \Rightarrow {\mathrm{n}}_2^2=\frac{\mathrm{R\lambda }}{\mathrm{R\lambda }-1} \\ \Rightarrow {\mathrm{n}}_2=\sqrt{\frac{\mathrm{\lambda R}}{\mathrm{\lambda R}-1}} \end{gathered}$$

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1+{\mathrm{E}}_2; \frac{\mathrm{hc}}{\mathrm{\lambda }}=\frac{\mathrm{hc}}{{\mathrm{\lambda }}_1}+\frac{\mathrm{hc}}{{\mathrm{\lambda }}_2} \\ \Rightarrow \frac{\mathrm{hc}}{\mathrm{\lambda }}\mathrm{hc} \left(\frac{{\mathrm{\lambda }}_2+{\mathrm{\lambda }}_1}{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}\right); \mathrm{\lambda }=2\frac{{\mathrm{\lambda }}_1{\mathrm{\lambda }}_2}{{\mathrm{\lambda }}_1+{\mathrm{\lambda }}_2} \end{gathered}$$

10 th element is neon, which is last element of 2nd period.