142457
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{\text {, is }}$
1 $\frac{\mathrm{h}}{\sqrt{\mathrm{mkT}}}$
2 $\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
3 $\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}$
4 $\frac{2 \mathrm{~h}}{\sqrt{\mathrm{mkT}}}$
Explanation:
B Energy of neutron $=$ K.E. per molecule of water $\mathrm{E}_{\mathrm{n}}=\frac{3}{2} \mathrm{kT}$ de-Brogile wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{\mathrm{n}}}}$ $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{3}{2} \mathrm{kT}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
NEET - 2017
Dual nature of radiation and Matter
142458
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ is Given: Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
142459
If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 \%$. Then the initial momentum of the electron is
1 $\frac{20}{\mathrm{P}}$
2 $20 \mathrm{P}$
3 $\frac{P}{20}$
4 $30 \mathrm{P}$
Explanation:
B We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|$ $\frac{\mathrm{p}}{\mathrm{p}_{\text {initial }}}=\frac{5}{100}$ $\mathrm{p}_{\text {initial }} \times 5=100 \mathrm{P}$ $\mathrm{p}_{\text {initial }}=20 \mathrm{P}$
Shift-II
Dual nature of radiation and Matter
142460
The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ respectively is
1 $2: 3$
2 $2 \sqrt{2}: \sqrt{3}$
3 $\sqrt{3}: 2 \sqrt{2}$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{He}}=127+273=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{H}_{2}}=27+273=300 \mathrm{~K}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}$ The ratio of de-Broglie wavelength of hydrogen and helium. $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \cdot \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \cdot \mathrm{T}_{\mathrm{H}_{2}}}}$ $=\sqrt{\left(\frac{4}{2}\right) \times \frac{400}{300}}$ $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{2}}{\sqrt{3}}$
Shift-II
Dual nature of radiation and Matter
142461
The wavelength $\lambda$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck constant.
1 $\frac{\lambda \mathrm{mc}}{\mathrm{h}}$
2 $\frac{\mathrm{hmc}}{\lambda}$
3 $\frac{2 \mathrm{hmc}}{\lambda}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Kinetic energy of electron $\mathrm{KE}_{\mathrm{e}}=\frac{\mathrm{mv}^{2}}{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE}_{\mathrm{e}}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ Energy of photon $\left(E_{\mathrm{P}}\right)=\frac{\mathrm{hc}}{\lambda}$ Ratio of kinetic energy of electron $\left(\mathrm{KE}_{\mathrm{e}}\right)$ and photon $\left(\mathrm{E}_{\mathrm{p}}\right)$ $\frac{\mathrm{KE}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda \mathrm{c}}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\left(\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}\right)$
142457
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{\text {, is }}$
1 $\frac{\mathrm{h}}{\sqrt{\mathrm{mkT}}}$
2 $\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
3 $\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}$
4 $\frac{2 \mathrm{~h}}{\sqrt{\mathrm{mkT}}}$
Explanation:
B Energy of neutron $=$ K.E. per molecule of water $\mathrm{E}_{\mathrm{n}}=\frac{3}{2} \mathrm{kT}$ de-Brogile wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{\mathrm{n}}}}$ $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{3}{2} \mathrm{kT}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
NEET - 2017
Dual nature of radiation and Matter
142458
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ is Given: Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
142459
If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 \%$. Then the initial momentum of the electron is
1 $\frac{20}{\mathrm{P}}$
2 $20 \mathrm{P}$
3 $\frac{P}{20}$
4 $30 \mathrm{P}$
Explanation:
B We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|$ $\frac{\mathrm{p}}{\mathrm{p}_{\text {initial }}}=\frac{5}{100}$ $\mathrm{p}_{\text {initial }} \times 5=100 \mathrm{P}$ $\mathrm{p}_{\text {initial }}=20 \mathrm{P}$
Shift-II
Dual nature of radiation and Matter
142460
The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ respectively is
1 $2: 3$
2 $2 \sqrt{2}: \sqrt{3}$
3 $\sqrt{3}: 2 \sqrt{2}$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{He}}=127+273=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{H}_{2}}=27+273=300 \mathrm{~K}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}$ The ratio of de-Broglie wavelength of hydrogen and helium. $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \cdot \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \cdot \mathrm{T}_{\mathrm{H}_{2}}}}$ $=\sqrt{\left(\frac{4}{2}\right) \times \frac{400}{300}}$ $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{2}}{\sqrt{3}}$
Shift-II
Dual nature of radiation and Matter
142461
The wavelength $\lambda$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck constant.
1 $\frac{\lambda \mathrm{mc}}{\mathrm{h}}$
2 $\frac{\mathrm{hmc}}{\lambda}$
3 $\frac{2 \mathrm{hmc}}{\lambda}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Kinetic energy of electron $\mathrm{KE}_{\mathrm{e}}=\frac{\mathrm{mv}^{2}}{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE}_{\mathrm{e}}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ Energy of photon $\left(E_{\mathrm{P}}\right)=\frac{\mathrm{hc}}{\lambda}$ Ratio of kinetic energy of electron $\left(\mathrm{KE}_{\mathrm{e}}\right)$ and photon $\left(\mathrm{E}_{\mathrm{p}}\right)$ $\frac{\mathrm{KE}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda \mathrm{c}}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\left(\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}\right)$
142457
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{\text {, is }}$
1 $\frac{\mathrm{h}}{\sqrt{\mathrm{mkT}}}$
2 $\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
3 $\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}$
4 $\frac{2 \mathrm{~h}}{\sqrt{\mathrm{mkT}}}$
Explanation:
B Energy of neutron $=$ K.E. per molecule of water $\mathrm{E}_{\mathrm{n}}=\frac{3}{2} \mathrm{kT}$ de-Brogile wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{\mathrm{n}}}}$ $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{3}{2} \mathrm{kT}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
NEET - 2017
Dual nature of radiation and Matter
142458
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ is Given: Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
142459
If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 \%$. Then the initial momentum of the electron is
1 $\frac{20}{\mathrm{P}}$
2 $20 \mathrm{P}$
3 $\frac{P}{20}$
4 $30 \mathrm{P}$
Explanation:
B We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|$ $\frac{\mathrm{p}}{\mathrm{p}_{\text {initial }}}=\frac{5}{100}$ $\mathrm{p}_{\text {initial }} \times 5=100 \mathrm{P}$ $\mathrm{p}_{\text {initial }}=20 \mathrm{P}$
Shift-II
Dual nature of radiation and Matter
142460
The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ respectively is
1 $2: 3$
2 $2 \sqrt{2}: \sqrt{3}$
3 $\sqrt{3}: 2 \sqrt{2}$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{He}}=127+273=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{H}_{2}}=27+273=300 \mathrm{~K}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}$ The ratio of de-Broglie wavelength of hydrogen and helium. $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \cdot \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \cdot \mathrm{T}_{\mathrm{H}_{2}}}}$ $=\sqrt{\left(\frac{4}{2}\right) \times \frac{400}{300}}$ $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{2}}{\sqrt{3}}$
Shift-II
Dual nature of radiation and Matter
142461
The wavelength $\lambda$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck constant.
1 $\frac{\lambda \mathrm{mc}}{\mathrm{h}}$
2 $\frac{\mathrm{hmc}}{\lambda}$
3 $\frac{2 \mathrm{hmc}}{\lambda}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Kinetic energy of electron $\mathrm{KE}_{\mathrm{e}}=\frac{\mathrm{mv}^{2}}{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE}_{\mathrm{e}}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ Energy of photon $\left(E_{\mathrm{P}}\right)=\frac{\mathrm{hc}}{\lambda}$ Ratio of kinetic energy of electron $\left(\mathrm{KE}_{\mathrm{e}}\right)$ and photon $\left(\mathrm{E}_{\mathrm{p}}\right)$ $\frac{\mathrm{KE}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda \mathrm{c}}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\left(\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}\right)$
142457
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{\text {, is }}$
1 $\frac{\mathrm{h}}{\sqrt{\mathrm{mkT}}}$
2 $\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
3 $\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}$
4 $\frac{2 \mathrm{~h}}{\sqrt{\mathrm{mkT}}}$
Explanation:
B Energy of neutron $=$ K.E. per molecule of water $\mathrm{E}_{\mathrm{n}}=\frac{3}{2} \mathrm{kT}$ de-Brogile wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{\mathrm{n}}}}$ $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{3}{2} \mathrm{kT}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
NEET - 2017
Dual nature of radiation and Matter
142458
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ is Given: Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
142459
If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 \%$. Then the initial momentum of the electron is
1 $\frac{20}{\mathrm{P}}$
2 $20 \mathrm{P}$
3 $\frac{P}{20}$
4 $30 \mathrm{P}$
Explanation:
B We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|$ $\frac{\mathrm{p}}{\mathrm{p}_{\text {initial }}}=\frac{5}{100}$ $\mathrm{p}_{\text {initial }} \times 5=100 \mathrm{P}$ $\mathrm{p}_{\text {initial }}=20 \mathrm{P}$
Shift-II
Dual nature of radiation and Matter
142460
The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ respectively is
1 $2: 3$
2 $2 \sqrt{2}: \sqrt{3}$
3 $\sqrt{3}: 2 \sqrt{2}$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{He}}=127+273=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{H}_{2}}=27+273=300 \mathrm{~K}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}$ The ratio of de-Broglie wavelength of hydrogen and helium. $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \cdot \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \cdot \mathrm{T}_{\mathrm{H}_{2}}}}$ $=\sqrt{\left(\frac{4}{2}\right) \times \frac{400}{300}}$ $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{2}}{\sqrt{3}}$
Shift-II
Dual nature of radiation and Matter
142461
The wavelength $\lambda$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck constant.
1 $\frac{\lambda \mathrm{mc}}{\mathrm{h}}$
2 $\frac{\mathrm{hmc}}{\lambda}$
3 $\frac{2 \mathrm{hmc}}{\lambda}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Kinetic energy of electron $\mathrm{KE}_{\mathrm{e}}=\frac{\mathrm{mv}^{2}}{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE}_{\mathrm{e}}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ Energy of photon $\left(E_{\mathrm{P}}\right)=\frac{\mathrm{hc}}{\lambda}$ Ratio of kinetic energy of electron $\left(\mathrm{KE}_{\mathrm{e}}\right)$ and photon $\left(\mathrm{E}_{\mathrm{p}}\right)$ $\frac{\mathrm{KE}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda \mathrm{c}}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\left(\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}\right)$
142457
The de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature $T$ (Kelvin) and mass $m_{\text {, is }}$
1 $\frac{\mathrm{h}}{\sqrt{\mathrm{mkT}}}$
2 $\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
3 $\frac{2 \mathrm{~h}}{\sqrt{3 \mathrm{mkT}}}$
4 $\frac{2 \mathrm{~h}}{\sqrt{\mathrm{mkT}}}$
Explanation:
B Energy of neutron $=$ K.E. per molecule of water $\mathrm{E}_{\mathrm{n}}=\frac{3}{2} \mathrm{kT}$ de-Brogile wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}_{\mathrm{n}}}}$ $=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{3}{2} \mathrm{kT}}}$ $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mkT}}}$
NEET - 2017
Dual nature of radiation and Matter
142458
The de-Broglie wavelength of a tennis ball of mass $60 \mathrm{~g}$ moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$ is Given: Planck's constant $=6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}$
142459
If the momentum of an electron changes by ' $P$ ' then the de-Broglie wavelength associated with it changes by $5 \%$. Then the initial momentum of the electron is
1 $\frac{20}{\mathrm{P}}$
2 $20 \mathrm{P}$
3 $\frac{P}{20}$
4 $30 \mathrm{P}$
Explanation:
B We know that, de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\left|\frac{\Delta \mathrm{p}}{\mathrm{p}}\right|=\left|\frac{\Delta \lambda}{\lambda}\right|$ $\frac{\mathrm{p}}{\mathrm{p}_{\text {initial }}}=\frac{5}{100}$ $\mathrm{p}_{\text {initial }} \times 5=100 \mathrm{P}$ $\mathrm{p}_{\text {initial }}=20 \mathrm{P}$
Shift-II
Dual nature of radiation and Matter
142460
The ratio of de-Broglie wavelength of molecules of hydrogen and helium which are at temperatures $27^{\circ} \mathrm{C}$ and $127^{\circ} \mathrm{C}$ respectively is
1 $2: 3$
2 $2 \sqrt{2}: \sqrt{3}$
3 $\sqrt{3}: 2 \sqrt{2}$
4 $\sqrt{2}: \sqrt{3}$
Explanation:
B Given that, $\mathrm{T}_{\mathrm{He}}=127+273=400 \mathrm{~K}$ $\mathrm{~T}_{\mathrm{H}_{2}}=27+273=300 \mathrm{~K}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{3 \mathrm{mKT}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}$ The ratio of de-Broglie wavelength of hydrogen and helium. $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \cdot \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \cdot \mathrm{T}_{\mathrm{H}_{2}}}}$ $=\sqrt{\left(\frac{4}{2}\right) \times \frac{400}{300}}$ $\frac{\lambda_{\mathrm{H}_{2}}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{8}{3}}=\frac{2 \sqrt{2}}{\sqrt{3}}$
Shift-II
Dual nature of radiation and Matter
142461
The wavelength $\lambda$ of a photon and the deBroglie wavelength of an electron have the same value. Find the ratio of energy of photon of the kinetic energy of electron in terms of mass $m$, speed of light $c$ and Planck constant.
1 $\frac{\lambda \mathrm{mc}}{\mathrm{h}}$
2 $\frac{\mathrm{hmc}}{\lambda}$
3 $\frac{2 \mathrm{hmc}}{\lambda}$
4 $\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}$
Explanation:
D de - Broglie wavelength $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Momentum, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Kinetic energy of electron $\mathrm{KE}_{\mathrm{e}}=\frac{\mathrm{mv}^{2}}{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ $\mathrm{KE}_{\mathrm{e}}=\frac{\left(\frac{\mathrm{h}}{\lambda}\right)^{2}}{2 \mathrm{~m}}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ Energy of photon $\left(E_{\mathrm{P}}\right)=\frac{\mathrm{hc}}{\lambda}$ Ratio of kinetic energy of electron $\left(\mathrm{KE}_{\mathrm{e}}\right)$ and photon $\left(\mathrm{E}_{\mathrm{p}}\right)$ $\frac{\mathrm{KE}_{\mathrm{e}}}{\mathrm{E}_{\mathrm{p}}}=\frac{\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}}{\frac{\mathrm{hc}}{\lambda}}=\frac{\mathrm{h}}{2 \mathrm{~m} \lambda \mathrm{c}}$ $\frac{\mathrm{E}_{\mathrm{p}}}{\mathrm{KE}_{\mathrm{e}}}=\left(\frac{2 \lambda \mathrm{mc}}{\mathrm{h}}\right)$
