142489
The distance of closest approach of an $\alpha$ particle fired towards a nucleus with momentum $p$, is $r$. If the momentum of the $\alpha$ particle is $2 \mathrm{p}$, the corresponding distance of closest approach is
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $4 \mathrm{r}$
4 $\frac{r}{8}$
5 $\frac{r}{4}$
Explanation:
E We know that, $\mathrm{p}_{1}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{1}}}$ $\mathrm{p}_{2}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{2}}}$ Equation (ii) divided by equation (i), we get - $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{k} / \sqrt{\mathrm{r}_{2}}}{\mathrm{k} / \sqrt{\mathrm{r}_{1}}}$ $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}$ Squaring on both side in equation (iii), we get- $\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}=\left(\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}\right)^{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}$ According to question $\mathrm{p}_{2} \rightarrow 2 \mathrm{p}$ $\mathrm{r}_{1}=\mathrm{r}$ Then from equation (iv) $\frac{r}{r_{2}}=\left(\frac{2 p}{p}\right)^{2}$ $\frac{r}{r_{2}}=4$ $r_{2}=\frac{r}{4}$
Kerala CEE - 2011
Dual nature of radiation and Matter
142490
A proton, a deuteron and an alpha particle with the same kinetic energy enter a region of uniform magnetic field $B$ at right angles to the field. The ratio of the radii of their circular paths is :
1 $1: 1: 1$
2 $1: \sqrt{2}: \sqrt{2}$
3 $\sqrt{2}: 1: 1$
4 $\sqrt{2}: \sqrt{2}: 1$
5 $1: \sqrt{2}: 1$
Explanation:
E Given, a proton, a deuteron and an alpha particle with the same kinetic energy. We know that, $\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}$ $\mathrm{R}=\frac{\mathrm{mv}^{2}}{\mathrm{qvB}}=\frac{\mathrm{mv}}{\mathrm{qB}}$ For proton, $R_{p}=\frac{m v}{q_{p} B}=\frac{\sqrt{2 m_{p} E}}{q_{p} B}$ For deuteron and $\alpha$-particle, $\mathrm{R}_{\mathrm{d}}=\frac{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{E}}}{\mathrm{q}_{\mathrm{d}} \mathrm{B}}$ and $\mathrm{R}_{\alpha}=\frac{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{E}}}{\mathrm{q}_{\alpha} \mathrm{B}}$ According to question, $\mathrm{R}_{\mathrm{p}}: \mathrm{R}_{\mathrm{d}}: \mathrm{R}_{\alpha}$ $\frac{\sqrt{\mathrm{m}_{\mathrm{p}}}}{\mathrm{q}_{\mathrm{p}}}: \frac{\sqrt{\mathrm{m}_{\mathrm{d}}}}{\mathrm{q}_{\mathrm{d}}}: \frac{\sqrt{\mathrm{m}_{\alpha}}}{\mathrm{q}_{\alpha}}$ $\therefore \frac{\sqrt{1}}{1}: \frac{\sqrt{2}}{1}: \frac{\sqrt{4}}{2}$ $\text { or } 1: \sqrt{2}: 1$
Kerala CEE 2006
Dual nature of radiation and Matter
142491
The ratio of the de-Broglie wavelength of an $\alpha$ particle and a proton of same kinetic energy is:
1 $1: 2$
2 $1: 1$
3 $1: \sqrt{2}$
4 $4: 1$
5 $\sqrt{2}: 1$
Explanation:
A From de - Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ For same kinetic energy, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\text { K.E. })}} \quad\left(\because \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times(\text { K.E. })}}: \frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m}_{\mathrm{p}} \times(\text { K.E. })}}$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=1: 2$
Kerala CEE 2005
Dual nature of radiation and Matter
142492
The wave length of a $1 \mathrm{keV}$ photon is $1.24 \mathrm{~nm}$. The frequency of $1 \mathrm{MeV}$ photon is:
142489
The distance of closest approach of an $\alpha$ particle fired towards a nucleus with momentum $p$, is $r$. If the momentum of the $\alpha$ particle is $2 \mathrm{p}$, the corresponding distance of closest approach is
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $4 \mathrm{r}$
4 $\frac{r}{8}$
5 $\frac{r}{4}$
Explanation:
E We know that, $\mathrm{p}_{1}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{1}}}$ $\mathrm{p}_{2}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{2}}}$ Equation (ii) divided by equation (i), we get - $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{k} / \sqrt{\mathrm{r}_{2}}}{\mathrm{k} / \sqrt{\mathrm{r}_{1}}}$ $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}$ Squaring on both side in equation (iii), we get- $\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}=\left(\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}\right)^{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}$ According to question $\mathrm{p}_{2} \rightarrow 2 \mathrm{p}$ $\mathrm{r}_{1}=\mathrm{r}$ Then from equation (iv) $\frac{r}{r_{2}}=\left(\frac{2 p}{p}\right)^{2}$ $\frac{r}{r_{2}}=4$ $r_{2}=\frac{r}{4}$
Kerala CEE - 2011
Dual nature of radiation and Matter
142490
A proton, a deuteron and an alpha particle with the same kinetic energy enter a region of uniform magnetic field $B$ at right angles to the field. The ratio of the radii of their circular paths is :
1 $1: 1: 1$
2 $1: \sqrt{2}: \sqrt{2}$
3 $\sqrt{2}: 1: 1$
4 $\sqrt{2}: \sqrt{2}: 1$
5 $1: \sqrt{2}: 1$
Explanation:
E Given, a proton, a deuteron and an alpha particle with the same kinetic energy. We know that, $\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}$ $\mathrm{R}=\frac{\mathrm{mv}^{2}}{\mathrm{qvB}}=\frac{\mathrm{mv}}{\mathrm{qB}}$ For proton, $R_{p}=\frac{m v}{q_{p} B}=\frac{\sqrt{2 m_{p} E}}{q_{p} B}$ For deuteron and $\alpha$-particle, $\mathrm{R}_{\mathrm{d}}=\frac{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{E}}}{\mathrm{q}_{\mathrm{d}} \mathrm{B}}$ and $\mathrm{R}_{\alpha}=\frac{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{E}}}{\mathrm{q}_{\alpha} \mathrm{B}}$ According to question, $\mathrm{R}_{\mathrm{p}}: \mathrm{R}_{\mathrm{d}}: \mathrm{R}_{\alpha}$ $\frac{\sqrt{\mathrm{m}_{\mathrm{p}}}}{\mathrm{q}_{\mathrm{p}}}: \frac{\sqrt{\mathrm{m}_{\mathrm{d}}}}{\mathrm{q}_{\mathrm{d}}}: \frac{\sqrt{\mathrm{m}_{\alpha}}}{\mathrm{q}_{\alpha}}$ $\therefore \frac{\sqrt{1}}{1}: \frac{\sqrt{2}}{1}: \frac{\sqrt{4}}{2}$ $\text { or } 1: \sqrt{2}: 1$
Kerala CEE 2006
Dual nature of radiation and Matter
142491
The ratio of the de-Broglie wavelength of an $\alpha$ particle and a proton of same kinetic energy is:
1 $1: 2$
2 $1: 1$
3 $1: \sqrt{2}$
4 $4: 1$
5 $\sqrt{2}: 1$
Explanation:
A From de - Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ For same kinetic energy, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\text { K.E. })}} \quad\left(\because \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times(\text { K.E. })}}: \frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m}_{\mathrm{p}} \times(\text { K.E. })}}$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=1: 2$
Kerala CEE 2005
Dual nature of radiation and Matter
142492
The wave length of a $1 \mathrm{keV}$ photon is $1.24 \mathrm{~nm}$. The frequency of $1 \mathrm{MeV}$ photon is:
142489
The distance of closest approach of an $\alpha$ particle fired towards a nucleus with momentum $p$, is $r$. If the momentum of the $\alpha$ particle is $2 \mathrm{p}$, the corresponding distance of closest approach is
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $4 \mathrm{r}$
4 $\frac{r}{8}$
5 $\frac{r}{4}$
Explanation:
E We know that, $\mathrm{p}_{1}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{1}}}$ $\mathrm{p}_{2}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{2}}}$ Equation (ii) divided by equation (i), we get - $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{k} / \sqrt{\mathrm{r}_{2}}}{\mathrm{k} / \sqrt{\mathrm{r}_{1}}}$ $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}$ Squaring on both side in equation (iii), we get- $\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}=\left(\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}\right)^{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}$ According to question $\mathrm{p}_{2} \rightarrow 2 \mathrm{p}$ $\mathrm{r}_{1}=\mathrm{r}$ Then from equation (iv) $\frac{r}{r_{2}}=\left(\frac{2 p}{p}\right)^{2}$ $\frac{r}{r_{2}}=4$ $r_{2}=\frac{r}{4}$
Kerala CEE - 2011
Dual nature of radiation and Matter
142490
A proton, a deuteron and an alpha particle with the same kinetic energy enter a region of uniform magnetic field $B$ at right angles to the field. The ratio of the radii of their circular paths is :
1 $1: 1: 1$
2 $1: \sqrt{2}: \sqrt{2}$
3 $\sqrt{2}: 1: 1$
4 $\sqrt{2}: \sqrt{2}: 1$
5 $1: \sqrt{2}: 1$
Explanation:
E Given, a proton, a deuteron and an alpha particle with the same kinetic energy. We know that, $\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}$ $\mathrm{R}=\frac{\mathrm{mv}^{2}}{\mathrm{qvB}}=\frac{\mathrm{mv}}{\mathrm{qB}}$ For proton, $R_{p}=\frac{m v}{q_{p} B}=\frac{\sqrt{2 m_{p} E}}{q_{p} B}$ For deuteron and $\alpha$-particle, $\mathrm{R}_{\mathrm{d}}=\frac{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{E}}}{\mathrm{q}_{\mathrm{d}} \mathrm{B}}$ and $\mathrm{R}_{\alpha}=\frac{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{E}}}{\mathrm{q}_{\alpha} \mathrm{B}}$ According to question, $\mathrm{R}_{\mathrm{p}}: \mathrm{R}_{\mathrm{d}}: \mathrm{R}_{\alpha}$ $\frac{\sqrt{\mathrm{m}_{\mathrm{p}}}}{\mathrm{q}_{\mathrm{p}}}: \frac{\sqrt{\mathrm{m}_{\mathrm{d}}}}{\mathrm{q}_{\mathrm{d}}}: \frac{\sqrt{\mathrm{m}_{\alpha}}}{\mathrm{q}_{\alpha}}$ $\therefore \frac{\sqrt{1}}{1}: \frac{\sqrt{2}}{1}: \frac{\sqrt{4}}{2}$ $\text { or } 1: \sqrt{2}: 1$
Kerala CEE 2006
Dual nature of radiation and Matter
142491
The ratio of the de-Broglie wavelength of an $\alpha$ particle and a proton of same kinetic energy is:
1 $1: 2$
2 $1: 1$
3 $1: \sqrt{2}$
4 $4: 1$
5 $\sqrt{2}: 1$
Explanation:
A From de - Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ For same kinetic energy, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\text { K.E. })}} \quad\left(\because \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times(\text { K.E. })}}: \frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m}_{\mathrm{p}} \times(\text { K.E. })}}$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=1: 2$
Kerala CEE 2005
Dual nature of radiation and Matter
142492
The wave length of a $1 \mathrm{keV}$ photon is $1.24 \mathrm{~nm}$. The frequency of $1 \mathrm{MeV}$ photon is:
142489
The distance of closest approach of an $\alpha$ particle fired towards a nucleus with momentum $p$, is $r$. If the momentum of the $\alpha$ particle is $2 \mathrm{p}$, the corresponding distance of closest approach is
1 $\frac{r}{2}$
2 $2 \mathrm{r}$
3 $4 \mathrm{r}$
4 $\frac{r}{8}$
5 $\frac{r}{4}$
Explanation:
E We know that, $\mathrm{p}_{1}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{1}}}$ $\mathrm{p}_{2}=\frac{\mathrm{k}}{\sqrt{\mathrm{r}_{2}}}$ Equation (ii) divided by equation (i), we get - $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\mathrm{k} / \sqrt{\mathrm{r}_{2}}}{\mathrm{k} / \sqrt{\mathrm{r}_{1}}}$ $\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}$ Squaring on both side in equation (iii), we get- $\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}=\left(\sqrt{\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}}\right)^{2}$ $\left(\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}\right)=\left(\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}\right)^{2}$ According to question $\mathrm{p}_{2} \rightarrow 2 \mathrm{p}$ $\mathrm{r}_{1}=\mathrm{r}$ Then from equation (iv) $\frac{r}{r_{2}}=\left(\frac{2 p}{p}\right)^{2}$ $\frac{r}{r_{2}}=4$ $r_{2}=\frac{r}{4}$
Kerala CEE - 2011
Dual nature of radiation and Matter
142490
A proton, a deuteron and an alpha particle with the same kinetic energy enter a region of uniform magnetic field $B$ at right angles to the field. The ratio of the radii of their circular paths is :
1 $1: 1: 1$
2 $1: \sqrt{2}: \sqrt{2}$
3 $\sqrt{2}: 1: 1$
4 $\sqrt{2}: \sqrt{2}: 1$
5 $1: \sqrt{2}: 1$
Explanation:
E Given, a proton, a deuteron and an alpha particle with the same kinetic energy. We know that, $\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{qvB}$ $\mathrm{R}=\frac{\mathrm{mv}^{2}}{\mathrm{qvB}}=\frac{\mathrm{mv}}{\mathrm{qB}}$ For proton, $R_{p}=\frac{m v}{q_{p} B}=\frac{\sqrt{2 m_{p} E}}{q_{p} B}$ For deuteron and $\alpha$-particle, $\mathrm{R}_{\mathrm{d}}=\frac{\sqrt{2 \mathrm{~m}_{\mathrm{d}} \mathrm{E}}}{\mathrm{q}_{\mathrm{d}} \mathrm{B}}$ and $\mathrm{R}_{\alpha}=\frac{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{E}}}{\mathrm{q}_{\alpha} \mathrm{B}}$ According to question, $\mathrm{R}_{\mathrm{p}}: \mathrm{R}_{\mathrm{d}}: \mathrm{R}_{\alpha}$ $\frac{\sqrt{\mathrm{m}_{\mathrm{p}}}}{\mathrm{q}_{\mathrm{p}}}: \frac{\sqrt{\mathrm{m}_{\mathrm{d}}}}{\mathrm{q}_{\mathrm{d}}}: \frac{\sqrt{\mathrm{m}_{\alpha}}}{\mathrm{q}_{\alpha}}$ $\therefore \frac{\sqrt{1}}{1}: \frac{\sqrt{2}}{1}: \frac{\sqrt{4}}{2}$ $\text { or } 1: \sqrt{2}: 1$
Kerala CEE 2006
Dual nature of radiation and Matter
142491
The ratio of the de-Broglie wavelength of an $\alpha$ particle and a proton of same kinetic energy is:
1 $1: 2$
2 $1: 1$
3 $1: \sqrt{2}$
4 $4: 1$
5 $\sqrt{2}: 1$
Explanation:
A From de - Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ For same kinetic energy, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\text { K.E. })}} \quad\left(\because \mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}\right)$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m}_{\mathrm{p}} \times(\text { K.E. })}}: \frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m}_{\mathrm{p}} \times(\text { K.E. })}}$ $\lambda_{\alpha}: \lambda_{\mathrm{p}}=1: 2$
Kerala CEE 2005
Dual nature of radiation and Matter
142492
The wave length of a $1 \mathrm{keV}$ photon is $1.24 \mathrm{~nm}$. The frequency of $1 \mathrm{MeV}$ photon is:
