142386
An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$ its deBroglie wavelength increases by $50 \%$. The value of $\left(V_{1} / V_{2}\right)$ is equal to
1 4
2 $9 / 4$
3 3
4 $3 / 2$
Explanation:
B From de-Broglie wavelength, $\lambda_{1}=\lambda, \lambda_{2}=\lambda+\lambda \times \frac{50}{100}=1.5 \lambda$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\frac{\lambda}{1.5 \lambda}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{2}{3}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{9}{4}$
Shift-II
Dual nature of radiation and Matter
142387
The de Broglie wavelength of a molecule in a gas at room temperature $(300 \mathrm{~K})$ is $\lambda_{1}$. If the temperature of the gas is increased to $600 \mathrm{~K}$, then the de Broglie wavelength of the same gas molecule becomes
142389
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its deBroglie wavelength will be:
1 $\frac{\lambda}{\sqrt{2}}$
2 $\frac{\lambda}{2}$
3 $2 \lambda$
4 $\sqrt{2} \lambda$
Explanation:
C Given that, Kinetic energy of electron $=\mathrm{E}$ Wavelength $=\lambda$ When kinetic energy of electron become $=\frac{E}{4}$ Then wavelength $\lambda^{\prime}=$ ? We know that, de-Brogli wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ When, $\mathrm{E} \rightarrow \frac{\mathrm{E}}{4}$ Then, $\quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}$ Equation (ii) dividing by equation (i), we get - $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}}{\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}}$ $\frac{\lambda^{\prime}}{\lambda}=2$ $\lambda^{\prime}=2 \lambda$
Shift-I
Dual nature of radiation and Matter
142390
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:
1 $4: 1$
2 $2: 1$
3 $8: 1$
4 $16: 1$
Explanation:
A We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}$ de-Brogli wavelength for proton, $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}}$ For $\alpha$ particle $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}_{\alpha}}}$ Equation (ii) divided by equation (i), $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\alpha} \mathrm{V}_{\alpha} \mathrm{q}_{\alpha}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=4: 1$
Shift-I
Dual nature of radiation and Matter
142391
The ratio of de-Broglie wavelength of an a particle and a proton accelerated form rest by the same potential is $\frac{1}{\sqrt{m}}$ the value of $m$ is
1 4
2 16
3 2
4 8
Explanation:
D From de Broglie wavelength is given by, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$ Here, $\mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}$ $\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}, \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{P}}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{8}}=\frac{1}{\sqrt{\mathrm{m}}}$ $\mathrm{m}=8$
142386
An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$ its deBroglie wavelength increases by $50 \%$. The value of $\left(V_{1} / V_{2}\right)$ is equal to
1 4
2 $9 / 4$
3 3
4 $3 / 2$
Explanation:
B From de-Broglie wavelength, $\lambda_{1}=\lambda, \lambda_{2}=\lambda+\lambda \times \frac{50}{100}=1.5 \lambda$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\frac{\lambda}{1.5 \lambda}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{2}{3}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{9}{4}$
Shift-II
Dual nature of radiation and Matter
142387
The de Broglie wavelength of a molecule in a gas at room temperature $(300 \mathrm{~K})$ is $\lambda_{1}$. If the temperature of the gas is increased to $600 \mathrm{~K}$, then the de Broglie wavelength of the same gas molecule becomes
142389
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its deBroglie wavelength will be:
1 $\frac{\lambda}{\sqrt{2}}$
2 $\frac{\lambda}{2}$
3 $2 \lambda$
4 $\sqrt{2} \lambda$
Explanation:
C Given that, Kinetic energy of electron $=\mathrm{E}$ Wavelength $=\lambda$ When kinetic energy of electron become $=\frac{E}{4}$ Then wavelength $\lambda^{\prime}=$ ? We know that, de-Brogli wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ When, $\mathrm{E} \rightarrow \frac{\mathrm{E}}{4}$ Then, $\quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}$ Equation (ii) dividing by equation (i), we get - $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}}{\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}}$ $\frac{\lambda^{\prime}}{\lambda}=2$ $\lambda^{\prime}=2 \lambda$
Shift-I
Dual nature of radiation and Matter
142390
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:
1 $4: 1$
2 $2: 1$
3 $8: 1$
4 $16: 1$
Explanation:
A We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}$ de-Brogli wavelength for proton, $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}}$ For $\alpha$ particle $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}_{\alpha}}}$ Equation (ii) divided by equation (i), $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\alpha} \mathrm{V}_{\alpha} \mathrm{q}_{\alpha}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=4: 1$
Shift-I
Dual nature of radiation and Matter
142391
The ratio of de-Broglie wavelength of an a particle and a proton accelerated form rest by the same potential is $\frac{1}{\sqrt{m}}$ the value of $m$ is
1 4
2 16
3 2
4 8
Explanation:
D From de Broglie wavelength is given by, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$ Here, $\mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}$ $\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}, \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{P}}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{8}}=\frac{1}{\sqrt{\mathrm{m}}}$ $\mathrm{m}=8$
142386
An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$ its deBroglie wavelength increases by $50 \%$. The value of $\left(V_{1} / V_{2}\right)$ is equal to
1 4
2 $9 / 4$
3 3
4 $3 / 2$
Explanation:
B From de-Broglie wavelength, $\lambda_{1}=\lambda, \lambda_{2}=\lambda+\lambda \times \frac{50}{100}=1.5 \lambda$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\frac{\lambda}{1.5 \lambda}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{2}{3}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{9}{4}$
Shift-II
Dual nature of radiation and Matter
142387
The de Broglie wavelength of a molecule in a gas at room temperature $(300 \mathrm{~K})$ is $\lambda_{1}$. If the temperature of the gas is increased to $600 \mathrm{~K}$, then the de Broglie wavelength of the same gas molecule becomes
142389
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its deBroglie wavelength will be:
1 $\frac{\lambda}{\sqrt{2}}$
2 $\frac{\lambda}{2}$
3 $2 \lambda$
4 $\sqrt{2} \lambda$
Explanation:
C Given that, Kinetic energy of electron $=\mathrm{E}$ Wavelength $=\lambda$ When kinetic energy of electron become $=\frac{E}{4}$ Then wavelength $\lambda^{\prime}=$ ? We know that, de-Brogli wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ When, $\mathrm{E} \rightarrow \frac{\mathrm{E}}{4}$ Then, $\quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}$ Equation (ii) dividing by equation (i), we get - $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}}{\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}}$ $\frac{\lambda^{\prime}}{\lambda}=2$ $\lambda^{\prime}=2 \lambda$
Shift-I
Dual nature of radiation and Matter
142390
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:
1 $4: 1$
2 $2: 1$
3 $8: 1$
4 $16: 1$
Explanation:
A We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}$ de-Brogli wavelength for proton, $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}}$ For $\alpha$ particle $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}_{\alpha}}}$ Equation (ii) divided by equation (i), $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\alpha} \mathrm{V}_{\alpha} \mathrm{q}_{\alpha}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=4: 1$
Shift-I
Dual nature of radiation and Matter
142391
The ratio of de-Broglie wavelength of an a particle and a proton accelerated form rest by the same potential is $\frac{1}{\sqrt{m}}$ the value of $m$ is
1 4
2 16
3 2
4 8
Explanation:
D From de Broglie wavelength is given by, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$ Here, $\mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}$ $\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}, \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{P}}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{8}}=\frac{1}{\sqrt{\mathrm{m}}}$ $\mathrm{m}=8$
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Dual nature of radiation and Matter
142386
An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$ its deBroglie wavelength increases by $50 \%$. The value of $\left(V_{1} / V_{2}\right)$ is equal to
1 4
2 $9 / 4$
3 3
4 $3 / 2$
Explanation:
B From de-Broglie wavelength, $\lambda_{1}=\lambda, \lambda_{2}=\lambda+\lambda \times \frac{50}{100}=1.5 \lambda$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\frac{\lambda}{1.5 \lambda}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{2}{3}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{9}{4}$
Shift-II
Dual nature of radiation and Matter
142387
The de Broglie wavelength of a molecule in a gas at room temperature $(300 \mathrm{~K})$ is $\lambda_{1}$. If the temperature of the gas is increased to $600 \mathrm{~K}$, then the de Broglie wavelength of the same gas molecule becomes
142389
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its deBroglie wavelength will be:
1 $\frac{\lambda}{\sqrt{2}}$
2 $\frac{\lambda}{2}$
3 $2 \lambda$
4 $\sqrt{2} \lambda$
Explanation:
C Given that, Kinetic energy of electron $=\mathrm{E}$ Wavelength $=\lambda$ When kinetic energy of electron become $=\frac{E}{4}$ Then wavelength $\lambda^{\prime}=$ ? We know that, de-Brogli wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ When, $\mathrm{E} \rightarrow \frac{\mathrm{E}}{4}$ Then, $\quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}$ Equation (ii) dividing by equation (i), we get - $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}}{\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}}$ $\frac{\lambda^{\prime}}{\lambda}=2$ $\lambda^{\prime}=2 \lambda$
Shift-I
Dual nature of radiation and Matter
142390
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:
1 $4: 1$
2 $2: 1$
3 $8: 1$
4 $16: 1$
Explanation:
A We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}$ de-Brogli wavelength for proton, $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}}$ For $\alpha$ particle $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}_{\alpha}}}$ Equation (ii) divided by equation (i), $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\alpha} \mathrm{V}_{\alpha} \mathrm{q}_{\alpha}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=4: 1$
Shift-I
Dual nature of radiation and Matter
142391
The ratio of de-Broglie wavelength of an a particle and a proton accelerated form rest by the same potential is $\frac{1}{\sqrt{m}}$ the value of $m$ is
1 4
2 16
3 2
4 8
Explanation:
D From de Broglie wavelength is given by, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$ Here, $\mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}$ $\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}, \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{P}}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{8}}=\frac{1}{\sqrt{\mathrm{m}}}$ $\mathrm{m}=8$
142386
An electron accelerated through a potential difference $V_{1}$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_{2}$ its deBroglie wavelength increases by $50 \%$. The value of $\left(V_{1} / V_{2}\right)$ is equal to
1 4
2 $9 / 4$
3 3
4 $3 / 2$
Explanation:
B From de-Broglie wavelength, $\lambda_{1}=\lambda, \lambda_{2}=\lambda+\lambda \times \frac{50}{100}=1.5 \lambda$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{V}}}$ $\frac{\lambda}{1.5 \lambda}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{2}{3}=\sqrt{\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}}$ $\frac{\mathrm{V}_{1}}{\mathrm{~V}_{2}}=\frac{9}{4}$
Shift-II
Dual nature of radiation and Matter
142387
The de Broglie wavelength of a molecule in a gas at room temperature $(300 \mathrm{~K})$ is $\lambda_{1}$. If the temperature of the gas is increased to $600 \mathrm{~K}$, then the de Broglie wavelength of the same gas molecule becomes
142389
The de Broglie wavelength of an electron having kinetic energy $E$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its deBroglie wavelength will be:
1 $\frac{\lambda}{\sqrt{2}}$
2 $\frac{\lambda}{2}$
3 $2 \lambda$
4 $\sqrt{2} \lambda$
Explanation:
C Given that, Kinetic energy of electron $=\mathrm{E}$ Wavelength $=\lambda$ When kinetic energy of electron become $=\frac{E}{4}$ Then wavelength $\lambda^{\prime}=$ ? We know that, de-Brogli wavelength, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ When, $\mathrm{E} \rightarrow \frac{\mathrm{E}}{4}$ Then, $\quad \lambda^{\prime}=\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}$ Equation (ii) dividing by equation (i), we get - $\frac{\lambda^{\prime}}{\lambda}=\frac{\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}}{\frac{\mathrm{h}}{\sqrt{2 \frac{\mathrm{mE}}{4}}}}$ $\frac{\lambda^{\prime}}{\lambda}=2$ $\lambda^{\prime}=2 \lambda$
Shift-I
Dual nature of radiation and Matter
142390
A proton and an $\alpha$-particle are accelerated from rest by $2 \mathrm{~V}$ and $4 \mathrm{~V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:
1 $4: 1$
2 $2: 1$
3 $8: 1$
4 $16: 1$
Explanation:
A We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mq} \Delta \mathrm{V}}}$ de-Brogli wavelength for proton, $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}}}}$ For $\alpha$ particle $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\alpha} \mathrm{q}_{\alpha} \mathrm{V}_{\alpha}}}$ Equation (ii) divided by equation (i), $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}} \mathrm{V}_{\mathrm{p}} \mathrm{q}_{\mathrm{p}}}{\mathrm{m}_{\alpha} \mathrm{V}_{\alpha} \mathrm{q}_{\alpha}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{1 \times 2 \times 1}{4 \times 4 \times 2}}=\frac{1}{4}$ $\lambda_{\mathrm{p}}: \lambda_{\alpha}=4: 1$
Shift-I
Dual nature of radiation and Matter
142391
The ratio of de-Broglie wavelength of an a particle and a proton accelerated form rest by the same potential is $\frac{1}{\sqrt{m}}$ the value of $m$ is
1 4
2 16
3 2
4 8
Explanation:
D From de Broglie wavelength is given by, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqv}}}$ Here, $\mathrm{m}_{\alpha}=4 \mathrm{~m}_{\mathrm{p}}$ $\mathrm{q}_{\alpha}=2 \mathrm{q}_{\mathrm{p}}$ $\lambda_{\alpha}=\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}, \lambda_{\mathrm{P}}=\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}$ $\frac{\lambda_{\alpha}}{\lambda_{\mathrm{P}}}=\frac{\frac{\mathrm{h}}{\sqrt{2 \times 4 \mathrm{~m} \times 2 \mathrm{eV}}}}{\frac{\mathrm{h}}{\sqrt{2 \times \mathrm{m} \times \mathrm{eV}}}}=\frac{1}{2 \sqrt{2}}=\frac{1}{\sqrt{8}}=\frac{1}{\sqrt{\mathrm{m}}}$ $\mathrm{m}=8$