Atomic Structure MCQs for NEET — Chemistry Questions with Answers

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.

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.