Please be patient as the PDF generation may take upto a minute.

Electrons of mass m with de- Broglie wavelength λ fall on the target in an X-ray tube. The cut-off wavelength (λ_o) of the emitted X-ray is:
1.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{mc\lambda }}^2}{\mathrm{h}}$
2.  ${\mathrm{\lambda }}_0=\frac{2\mathrm{h}}{\mathrm{mc}}$
3.  ${\mathrm{\lambda }}_0=\frac{2{\mathrm{m}}^2{\mathrm{c}}^2{\mathrm{\lambda }}^2}{{\mathrm{h}}^2}$
4.  ${\mathrm{\lambda }}_0=\mathrm{\lambda }$

Photons with energy \(5~\text{eV}\) are incident on a cathode \(C\) in a photoelectric cell. The maximum energy of emitted photoelectrons is \(2~\text{eV}.\) When photons of energy \(6~\text{eV}\) are incident on \(C,\) no photoelectrons will reach the anode \(A,\) if the stopping potential of \(A\) relative to \(C\) is:
1. \(+3~\text{V}\)
2. \(+4~\text{V}\)
3. \(-1~\text{V}\)
4. \(-3~\text{V}\)

An electron of mass \(m\) with an initial velocity \(\overrightarrow v= v_0\hat i\)$\stackrel{}{}$\( ( v_o > 0 ) \) enters in an electric field \(\overrightarrow E = -E_0 \hat i\) \((E_0 = \text{constant}>0)\) at \(t=0.\) If \(\lambda_0,\)${\mathrm{}}_{}$ is its de-Broglie wavelength initially, then what will be its de-Broglie wavelength at time \(t?\)
1. \(\frac{\lambda_0}{\left(1+ \frac{eE_0}{mv_0}t\right)}\)
2. \(\lambda_0\left(1+ \frac{eE_0}{mv_0}t\right)\)
3. \(\lambda_0 t\)
4. \(\lambda_0\)

The hydrogen gas with its atoms in the ground state is excited by monochromatic radiation of \(\lambda = 975~\mathring{{A}}.\) The number of spectral lines in the resulting spectrum emitted will be:
1. \(3\)
2. \(2\)
3. \(6\)
4. \(10\)

Due to transitions among its first three energy levels, the hydrogen atom emits radiation at three discrete wavelengths ${\mathrm{\lambda }}_1, {\mathrm{\lambda }}_2 \mathrm{and} {\mathrm{\lambda }}_3\left({\mathrm{\lambda }}_1 < {\mathrm{\lambda }}_2 < {\mathrm{\lambda }}_3\right)$. Then,
1.  ${\lambda }_1={\lambda }_2+{\lambda }_3$
2.  ${\lambda }_1+{\lambda }_2={\lambda }_3$
3.  $\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$
4.  $\frac{1}{{\lambda }_1}=\frac{1}{{\lambda }_2}+\frac{1}{{\lambda }_3}$

In a hydrogen atom, the electron is in nth excited state. Knowing that on its way down to the second excited state, the number of different wavelengths that the electron can possibly emit is 10  . What is the value of n? 
1. 6
2. 7
3. 8
4. 5

The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between 40% decay and 85% decay of the same radioactive substance is:
1. 15
2. 30
3. 45
4. 60

The Binding energy per nucleon of \(^{7}_{3}\mathrm{Li}\) and \(^{4}_{2}\mathrm{He}\) nucleon are \(5.60~\text{MeV}\) and \(7.06~\text{MeV}\), respectively. In the nuclear reaction \(^{7}_{3}\mathrm{Li} + ^{1}_{1}\mathrm{H} \rightarrow ^{4}_{2}\mathrm{He} + ^{4}_{2}\mathrm{He} +Q\), the value of energy \(Q\) released is:
1. \(19.6~\text{MeV}\)
 2. \(-2.4~\text{MeV}\)
 3. \(8.4~\text{MeV}\)
4. \(17.3~\text{MeV}\)

A nucleus \({ }_{{n}}^{{m}} \mathrm{X}\) emits one $\mathrm{}$\(\alpha\text -\text{particle}\) and two \(\beta\text- \text{particle}\) The resulting nucleus is:

The binding energy of deuteron is \(2.2~\text{MeV}\) and that of \(_2\mathrm{He}^{4}\) is \(28~\text{MeV}\). If two deuterons are fused to form one \(_{2}\mathrm{He}^{4}\), ${\mathrm{}}^{}$then the energy released is:
1. \(25.8~\text{MeV}\)
2. \(23.6~\text{MeV}\)
3. \(19.2~\text{MeV}\)
4. \(30.2~\text{MeV}\)

In a radioactive material, the activity at time t₁ is R₁ and at a later time t₂, it is R₂. If the decay constant of the material is λ, then:
1. $R_1=R_2{e}^{\lambda (t_1\mathit{+}{t}_2)}$
2. $R_1=R_2{e}^{-\lambda (t_1-t_2)}$
3. $R_1=R_2(t_1-t_2)$
4. $R_1=R_2$