QBManager

Dual nature of radiation and Matter

142435 A proton and an $α$ -particle are accelerated from rest to the same energy. The de-Broglie wavelength $λ_{p}$ and $λ_{α}$ are in the ratio

1

2 : 1

2

1 : 1

3

\sqrt{2} : 1

4

4 : 1

Explanation:

A We know that, wavelength de-Broglie wavelength-
$(λ) = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ \propto \frac{1}{\sqrt{m}}$
$\sqrt{\frac{m_{α}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\sqrt{\frac{4 m_{p}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\frac{λ_{p}}{λ_{α}} = \frac{2}{1}$

NEET (Odisha)- 2019

Dual nature of radiation and Matter

142437 The ratio of de Broglie wavelengths of molecules of Hydrogen and Helium in two different gas jars at temperatures of $127^{\circ} C$ and $227^{\circ} C$ respectively is

1

\sqrt{\frac{2}{5}}

2

\sqrt{\frac{5}{2}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{8}{5}}

Explanation:

B Given, }\mathrm{T}_{\mathrm{H}_{2}}=127+273=400 \mathrm{~K}\mathrm{~T}_{\mathrm{He}}=227+273=500 \mathrm{~K} $W e k n o w t h a t,$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKT}}}\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \mathrm{~T}_{\mathrm{H}_{2}}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{4 \times 500}{2 \times 400}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{5}{2}}$

Shift-II

Dual nature of radiation and Matter

142438 A particle performs uniform circular motion with an angular momentum $L$ . If the frequency of particle's motion is doubled and its kinetic energy is halved, then the angular momentum becomes

1

\frac{L}{2}

2

\frac{L}{4}

3

\frac{L}{6}

4

\frac{L}{8}

Explanation:

B Rotational kinetic energy
$KE = \frac{1}{2} I ω^{2}$
$I = \frac{2 KE}{ω^{2}}$
Angular momentum $L = I ω$
$L = \frac{2 KE}{ω^{2}} \cdot ω = \frac{2 KE}{ω}$
${KE}_{1} = \frac{KE}{2} and ω_{1} = 2 ω$
$L^{'} = \frac{2 {KE}_{1}}{ω_{1}} = \frac{2 (\frac{KE}{2})}{2 ω}$
$L^{'} = \frac{1}{4} \frac{2 KE}{ω} = \frac{L}{4}$

Assam CEE-2019

Dual nature of radiation and Matter

142439 A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(λ_{p})$ and that of electron $(λ_{e})$ ?

1

λ_{p} = 2000 λ_{e}

2

λ_{p} = \frac{λ_{e}}{2000}

3

λ_{p} = 20 \sqrt{5} λ_{e}

4

λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}

Explanation:

D Given,
Mass of proton $(m_{p}) = 2000$ mass of electron $(m_{e})$
$\frac{λ_{p}}{λ_{e}} = ?$
$K = \frac{1}{2} {mv}^{2} \times \frac{m}{m}$
$K = \frac{1}{2} \frac{(mv)^{2}}{m} = \frac{p^{2}}{2 m}$
$p^{2} = 2 mK$
$p = \sqrt{2 mK}$
de Broglie wavelength-
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}$
So, $λ_{p} = \frac{h}{\sqrt{2 m_{p} eV}}$
$K = eV$
$V_{p} = V_{e}$
And, $λ_{e} = \frac{h}{\sqrt{2 m_{e} eV}}$
Now, $\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} = \frac{\sqrt{m_{e}}}{\sqrt{2000 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{e}}} \times \frac{1}{\sqrt{2000}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{20 \sqrt{5}}$
Hence, $λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}$

WB JEE 2019

Dual nature of radiation and Matter

142440 What is the ratio of their final velocity when an alpha particle and a proton are accelerated at the same potential difference from the rest?

1

2 : 1

2

1 : 1

3

1 : \sqrt{2}

4

1 : 2

Explanation:

C We know that
$V = \frac{W}{q}$
$qV = \frac{1}{2} {mv}^{2}$
$v^{2} = \frac{2 qV}{m}$
For proton, $v_{P}^{2} = \frac{2 q_{P} V}{m_{P}}$
And for $α$ particle, $v_{α}^{2} = \frac{2 q_{α} V}{m_{α}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{q_{α}}{m_{α}} \cdot \frac{m_{P}}{q_{P}}$
The mass of $α$ -particle is 4 times the mass of proton and the charge of $α$ - particle is to times of proton then,
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{2 q_{P}}{4 m_{P}} \times \frac{m_{P}}{q_{P}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{1}{2}$
$\frac{v_{α}}{v_{P}} = \frac{1}{\sqrt{2}}$
$Or, v_{α} : v_{p} = 1 : \sqrt{2}$

J and K CET-2019

Dual nature of radiation and Matter

142435 A proton and an $α$ -particle are accelerated from rest to the same energy. The de-Broglie wavelength $λ_{p}$ and $λ_{α}$ are in the ratio

1

2 : 1

2

1 : 1

3

\sqrt{2} : 1

4

4 : 1

Explanation:

A We know that, wavelength de-Broglie wavelength-
$(λ) = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ \propto \frac{1}{\sqrt{m}}$
$\sqrt{\frac{m_{α}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\sqrt{\frac{4 m_{p}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\frac{λ_{p}}{λ_{α}} = \frac{2}{1}$

NEET (Odisha)- 2019

Dual nature of radiation and Matter

142437 The ratio of de Broglie wavelengths of molecules of Hydrogen and Helium in two different gas jars at temperatures of $127^{\circ} C$ and $227^{\circ} C$ respectively is

1

\sqrt{\frac{2}{5}}

2

\sqrt{\frac{5}{2}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{8}{5}}

Explanation:

B Given, }\mathrm{T}_{\mathrm{H}_{2}}=127+273=400 \mathrm{~K}\mathrm{~T}_{\mathrm{He}}=227+273=500 \mathrm{~K} $W e k n o w t h a t,$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKT}}}\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \mathrm{~T}_{\mathrm{H}_{2}}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{4 \times 500}{2 \times 400}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{5}{2}}$

Shift-II

Dual nature of radiation and Matter

142438 A particle performs uniform circular motion with an angular momentum $L$ . If the frequency of particle's motion is doubled and its kinetic energy is halved, then the angular momentum becomes

1

\frac{L}{2}

2

\frac{L}{4}

3

\frac{L}{6}

4

\frac{L}{8}

Explanation:

B Rotational kinetic energy
$KE = \frac{1}{2} I ω^{2}$
$I = \frac{2 KE}{ω^{2}}$
Angular momentum $L = I ω$
$L = \frac{2 KE}{ω^{2}} \cdot ω = \frac{2 KE}{ω}$
${KE}_{1} = \frac{KE}{2} and ω_{1} = 2 ω$
$L^{'} = \frac{2 {KE}_{1}}{ω_{1}} = \frac{2 (\frac{KE}{2})}{2 ω}$
$L^{'} = \frac{1}{4} \frac{2 KE}{ω} = \frac{L}{4}$

Assam CEE-2019

Dual nature of radiation and Matter

142439 A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(λ_{p})$ and that of electron $(λ_{e})$ ?

1

λ_{p} = 2000 λ_{e}

2

λ_{p} = \frac{λ_{e}}{2000}

3

λ_{p} = 20 \sqrt{5} λ_{e}

4

λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}

Explanation:

D Given,
Mass of proton $(m_{p}) = 2000$ mass of electron $(m_{e})$
$\frac{λ_{p}}{λ_{e}} = ?$
$K = \frac{1}{2} {mv}^{2} \times \frac{m}{m}$
$K = \frac{1}{2} \frac{(mv)^{2}}{m} = \frac{p^{2}}{2 m}$
$p^{2} = 2 mK$
$p = \sqrt{2 mK}$
de Broglie wavelength-
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}$
So, $λ_{p} = \frac{h}{\sqrt{2 m_{p} eV}}$
$K = eV$
$V_{p} = V_{e}$
And, $λ_{e} = \frac{h}{\sqrt{2 m_{e} eV}}$
Now, $\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} = \frac{\sqrt{m_{e}}}{\sqrt{2000 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{e}}} \times \frac{1}{\sqrt{2000}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{20 \sqrt{5}}$
Hence, $λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}$

WB JEE 2019

Dual nature of radiation and Matter

142440 What is the ratio of their final velocity when an alpha particle and a proton are accelerated at the same potential difference from the rest?

1

2 : 1

2

1 : 1

3

1 : \sqrt{2}

4

1 : 2

Explanation:

C We know that
$V = \frac{W}{q}$
$qV = \frac{1}{2} {mv}^{2}$
$v^{2} = \frac{2 qV}{m}$
For proton, $v_{P}^{2} = \frac{2 q_{P} V}{m_{P}}$
And for $α$ particle, $v_{α}^{2} = \frac{2 q_{α} V}{m_{α}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{q_{α}}{m_{α}} \cdot \frac{m_{P}}{q_{P}}$
The mass of $α$ -particle is 4 times the mass of proton and the charge of $α$ - particle is to times of proton then,
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{2 q_{P}}{4 m_{P}} \times \frac{m_{P}}{q_{P}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{1}{2}$
$\frac{v_{α}}{v_{P}} = \frac{1}{\sqrt{2}}$
$Or, v_{α} : v_{p} = 1 : \sqrt{2}$

J and K CET-2019

Dual nature of radiation and Matter

142435 A proton and an $α$ -particle are accelerated from rest to the same energy. The de-Broglie wavelength $λ_{p}$ and $λ_{α}$ are in the ratio

1

2 : 1

2

1 : 1

3

\sqrt{2} : 1

4

4 : 1

Explanation:

A We know that, wavelength de-Broglie wavelength-
$(λ) = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ \propto \frac{1}{\sqrt{m}}$
$\sqrt{\frac{m_{α}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\sqrt{\frac{4 m_{p}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\frac{λ_{p}}{λ_{α}} = \frac{2}{1}$

NEET (Odisha)- 2019

Dual nature of radiation and Matter

142437 The ratio of de Broglie wavelengths of molecules of Hydrogen and Helium in two different gas jars at temperatures of $127^{\circ} C$ and $227^{\circ} C$ respectively is

1

\sqrt{\frac{2}{5}}

2

\sqrt{\frac{5}{2}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{8}{5}}

Explanation:

B Given, }\mathrm{T}_{\mathrm{H}_{2}}=127+273=400 \mathrm{~K}\mathrm{~T}_{\mathrm{He}}=227+273=500 \mathrm{~K} $W e k n o w t h a t,$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKT}}}\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \mathrm{~T}_{\mathrm{H}_{2}}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{4 \times 500}{2 \times 400}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{5}{2}}$

Shift-II

Dual nature of radiation and Matter

142438 A particle performs uniform circular motion with an angular momentum $L$ . If the frequency of particle's motion is doubled and its kinetic energy is halved, then the angular momentum becomes

1

\frac{L}{2}

2

\frac{L}{4}

3

\frac{L}{6}

4

\frac{L}{8}

Explanation:

B Rotational kinetic energy
$KE = \frac{1}{2} I ω^{2}$
$I = \frac{2 KE}{ω^{2}}$
Angular momentum $L = I ω$
$L = \frac{2 KE}{ω^{2}} \cdot ω = \frac{2 KE}{ω}$
${KE}_{1} = \frac{KE}{2} and ω_{1} = 2 ω$
$L^{'} = \frac{2 {KE}_{1}}{ω_{1}} = \frac{2 (\frac{KE}{2})}{2 ω}$
$L^{'} = \frac{1}{4} \frac{2 KE}{ω} = \frac{L}{4}$

Assam CEE-2019

Dual nature of radiation and Matter

142439 A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(λ_{p})$ and that of electron $(λ_{e})$ ?

1

λ_{p} = 2000 λ_{e}

2

λ_{p} = \frac{λ_{e}}{2000}

3

λ_{p} = 20 \sqrt{5} λ_{e}

4

λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}

Explanation:

D Given,
Mass of proton $(m_{p}) = 2000$ mass of electron $(m_{e})$
$\frac{λ_{p}}{λ_{e}} = ?$
$K = \frac{1}{2} {mv}^{2} \times \frac{m}{m}$
$K = \frac{1}{2} \frac{(mv)^{2}}{m} = \frac{p^{2}}{2 m}$
$p^{2} = 2 mK$
$p = \sqrt{2 mK}$
de Broglie wavelength-
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}$
So, $λ_{p} = \frac{h}{\sqrt{2 m_{p} eV}}$
$K = eV$
$V_{p} = V_{e}$
And, $λ_{e} = \frac{h}{\sqrt{2 m_{e} eV}}$
Now, $\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} = \frac{\sqrt{m_{e}}}{\sqrt{2000 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{e}}} \times \frac{1}{\sqrt{2000}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{20 \sqrt{5}}$
Hence, $λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}$

WB JEE 2019

Dual nature of radiation and Matter

142440 What is the ratio of their final velocity when an alpha particle and a proton are accelerated at the same potential difference from the rest?

1

2 : 1

2

1 : 1

3

1 : \sqrt{2}

4

1 : 2

Explanation:

C We know that
$V = \frac{W}{q}$
$qV = \frac{1}{2} {mv}^{2}$
$v^{2} = \frac{2 qV}{m}$
For proton, $v_{P}^{2} = \frac{2 q_{P} V}{m_{P}}$
And for $α$ particle, $v_{α}^{2} = \frac{2 q_{α} V}{m_{α}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{q_{α}}{m_{α}} \cdot \frac{m_{P}}{q_{P}}$
The mass of $α$ -particle is 4 times the mass of proton and the charge of $α$ - particle is to times of proton then,
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{2 q_{P}}{4 m_{P}} \times \frac{m_{P}}{q_{P}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{1}{2}$
$\frac{v_{α}}{v_{P}} = \frac{1}{\sqrt{2}}$
$Or, v_{α} : v_{p} = 1 : \sqrt{2}$

J and K CET-2019

Dual nature of radiation and Matter

142435 A proton and an $α$ -particle are accelerated from rest to the same energy. The de-Broglie wavelength $λ_{p}$ and $λ_{α}$ are in the ratio

1

2 : 1

2

1 : 1

3

\sqrt{2} : 1

4

4 : 1

Explanation:

A We know that, wavelength de-Broglie wavelength-
$(λ) = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ \propto \frac{1}{\sqrt{m}}$
$\sqrt{\frac{m_{α}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\sqrt{\frac{4 m_{p}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\frac{λ_{p}}{λ_{α}} = \frac{2}{1}$

NEET (Odisha)- 2019

Dual nature of radiation and Matter

142437 The ratio of de Broglie wavelengths of molecules of Hydrogen and Helium in two different gas jars at temperatures of $127^{\circ} C$ and $227^{\circ} C$ respectively is

1

\sqrt{\frac{2}{5}}

2

\sqrt{\frac{5}{2}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{8}{5}}

Explanation:

B Given, }\mathrm{T}_{\mathrm{H}_{2}}=127+273=400 \mathrm{~K}\mathrm{~T}_{\mathrm{He}}=227+273=500 \mathrm{~K} $W e k n o w t h a t,$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKT}}}\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \mathrm{~T}_{\mathrm{H}_{2}}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{4 \times 500}{2 \times 400}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{5}{2}}$

Shift-II

Dual nature of radiation and Matter

142438 A particle performs uniform circular motion with an angular momentum $L$ . If the frequency of particle's motion is doubled and its kinetic energy is halved, then the angular momentum becomes

1

\frac{L}{2}

2

\frac{L}{4}

3

\frac{L}{6}

4

\frac{L}{8}

Explanation:

B Rotational kinetic energy
$KE = \frac{1}{2} I ω^{2}$
$I = \frac{2 KE}{ω^{2}}$
Angular momentum $L = I ω$
$L = \frac{2 KE}{ω^{2}} \cdot ω = \frac{2 KE}{ω}$
${KE}_{1} = \frac{KE}{2} and ω_{1} = 2 ω$
$L^{'} = \frac{2 {KE}_{1}}{ω_{1}} = \frac{2 (\frac{KE}{2})}{2 ω}$
$L^{'} = \frac{1}{4} \frac{2 KE}{ω} = \frac{L}{4}$

Assam CEE-2019

Dual nature of radiation and Matter

142439 A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(λ_{p})$ and that of electron $(λ_{e})$ ?

1

λ_{p} = 2000 λ_{e}

2

λ_{p} = \frac{λ_{e}}{2000}

3

λ_{p} = 20 \sqrt{5} λ_{e}

4

λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}

Explanation:

D Given,
Mass of proton $(m_{p}) = 2000$ mass of electron $(m_{e})$
$\frac{λ_{p}}{λ_{e}} = ?$
$K = \frac{1}{2} {mv}^{2} \times \frac{m}{m}$
$K = \frac{1}{2} \frac{(mv)^{2}}{m} = \frac{p^{2}}{2 m}$
$p^{2} = 2 mK$
$p = \sqrt{2 mK}$
de Broglie wavelength-
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}$
So, $λ_{p} = \frac{h}{\sqrt{2 m_{p} eV}}$
$K = eV$
$V_{p} = V_{e}$
And, $λ_{e} = \frac{h}{\sqrt{2 m_{e} eV}}$
Now, $\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} = \frac{\sqrt{m_{e}}}{\sqrt{2000 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{e}}} \times \frac{1}{\sqrt{2000}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{20 \sqrt{5}}$
Hence, $λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}$

WB JEE 2019

Dual nature of radiation and Matter

142440 What is the ratio of their final velocity when an alpha particle and a proton are accelerated at the same potential difference from the rest?

1

2 : 1

2

1 : 1

3

1 : \sqrt{2}

4

1 : 2

Explanation:

C We know that
$V = \frac{W}{q}$
$qV = \frac{1}{2} {mv}^{2}$
$v^{2} = \frac{2 qV}{m}$
For proton, $v_{P}^{2} = \frac{2 q_{P} V}{m_{P}}$
And for $α$ particle, $v_{α}^{2} = \frac{2 q_{α} V}{m_{α}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{q_{α}}{m_{α}} \cdot \frac{m_{P}}{q_{P}}$
The mass of $α$ -particle is 4 times the mass of proton and the charge of $α$ - particle is to times of proton then,
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{2 q_{P}}{4 m_{P}} \times \frac{m_{P}}{q_{P}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{1}{2}$
$\frac{v_{α}}{v_{P}} = \frac{1}{\sqrt{2}}$
$Or, v_{α} : v_{p} = 1 : \sqrt{2}$

J and K CET-2019

Dual nature of radiation and Matter

142435 A proton and an $α$ -particle are accelerated from rest to the same energy. The de-Broglie wavelength $λ_{p}$ and $λ_{α}$ are in the ratio

1

2 : 1

2

1 : 1

3

\sqrt{2} : 1

4

4 : 1

Explanation:

A We know that, wavelength de-Broglie wavelength-
$(λ) = \frac{h}{p} = \frac{h}{\sqrt{2 m (KE)}}$
$λ \propto \frac{1}{\sqrt{m}}$
$\sqrt{\frac{m_{α}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\sqrt{\frac{4 m_{p}}{m_{p}}} = \frac{λ_{p}}{λ_{α}}$
$\frac{λ_{p}}{λ_{α}} = \frac{2}{1}$

NEET (Odisha)- 2019

Dual nature of radiation and Matter

142437 The ratio of de Broglie wavelengths of molecules of Hydrogen and Helium in two different gas jars at temperatures of $127^{\circ} C$ and $227^{\circ} C$ respectively is

1

\sqrt{\frac{2}{5}}

2

\sqrt{\frac{5}{2}}

3

\sqrt{\frac{3}{5}}

4

\sqrt{\frac{8}{5}}

Explanation:

B Given, }\mathrm{T}_{\mathrm{H}_{2}}=127+273=400 \mathrm{~K}\mathrm{~T}_{\mathrm{He}}=227+273=500 \mathrm{~K} $W e k n o w t h a t,$ \lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mKT}}}\lambda \propto \frac{1}{\sqrt{\mathrm{mT}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{He}} \mathrm{T}_{\mathrm{He}}}{\mathrm{m}_{\mathrm{H}_{2}} \mathrm{~T}_{\mathrm{H}_{2}}}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{4 \times 500}{2 \times 400}}\frac{\lambda_{\left(\mathrm{H}_{2}\right)}}{\lambda_{\mathrm{He}}}=\sqrt{\frac{5}{2}}$

Shift-II

Dual nature of radiation and Matter

142438 A particle performs uniform circular motion with an angular momentum $L$ . If the frequency of particle's motion is doubled and its kinetic energy is halved, then the angular momentum becomes

1

\frac{L}{2}

2

\frac{L}{4}

3

\frac{L}{6}

4

\frac{L}{8}

Explanation:

B Rotational kinetic energy
$KE = \frac{1}{2} I ω^{2}$
$I = \frac{2 KE}{ω^{2}}$
Angular momentum $L = I ω$
$L = \frac{2 KE}{ω^{2}} \cdot ω = \frac{2 KE}{ω}$
${KE}_{1} = \frac{KE}{2} and ω_{1} = 2 ω$
$L^{'} = \frac{2 {KE}_{1}}{ω_{1}} = \frac{2 (\frac{KE}{2})}{2 ω}$
$L^{'} = \frac{1}{4} \frac{2 KE}{ω} = \frac{L}{4}$

Assam CEE-2019

Dual nature of radiation and Matter

142439 A proton and an electron initially at rest are accelerated by the same potential difference. Assuming that a proton is 2000 times heavier than an electron, what will be the relation between the de Broglie wavelength of the proton $(λ_{p})$ and that of electron $(λ_{e})$ ?

1

λ_{p} = 2000 λ_{e}

2

λ_{p} = \frac{λ_{e}}{2000}

3

λ_{p} = 20 \sqrt{5} λ_{e}

4

λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}

Explanation:

D Given,
Mass of proton $(m_{p}) = 2000$ mass of electron $(m_{e})$
$\frac{λ_{p}}{λ_{e}} = ?$
$K = \frac{1}{2} {mv}^{2} \times \frac{m}{m}$
$K = \frac{1}{2} \frac{(mv)^{2}}{m} = \frac{p^{2}}{2 m}$
$p^{2} = 2 mK$
$p = \sqrt{2 mK}$
de Broglie wavelength-
$λ = \frac{h}{p} = \frac{h}{\sqrt{2 mK}}$
So, $λ_{p} = \frac{h}{\sqrt{2 m_{p} eV}}$
$K = eV$
$V_{p} = V_{e}$
And, $λ_{e} = \frac{h}{\sqrt{2 m_{e} eV}}$
Now, $\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{p}}} = \frac{\sqrt{m_{e}}}{\sqrt{2000 m_{e}}}$
$\frac{λ_{p}}{λ_{e}} = \frac{\sqrt{m_{e}}}{\sqrt{m_{e}}} \times \frac{1}{\sqrt{2000}}$
$\frac{λ_{p}}{λ_{e}} = \frac{1}{20 \sqrt{5}}$
Hence, $λ_{p} = \frac{λ_{e}}{20 \sqrt{5}}$

WB JEE 2019

Dual nature of radiation and Matter

142440 What is the ratio of their final velocity when an alpha particle and a proton are accelerated at the same potential difference from the rest?

1

2 : 1

2

1 : 1

3

1 : \sqrt{2}

4

1 : 2

Explanation:

C We know that
$V = \frac{W}{q}$
$qV = \frac{1}{2} {mv}^{2}$
$v^{2} = \frac{2 qV}{m}$
For proton, $v_{P}^{2} = \frac{2 q_{P} V}{m_{P}}$
And for $α$ particle, $v_{α}^{2} = \frac{2 q_{α} V}{m_{α}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{q_{α}}{m_{α}} \cdot \frac{m_{P}}{q_{P}}$
The mass of $α$ -particle is 4 times the mass of proton and the charge of $α$ - particle is to times of proton then,
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{2 q_{P}}{4 m_{P}} \times \frac{m_{P}}{q_{P}}$
$\frac{v_{α}^{2}}{v_{P}^{2}} = \frac{1}{2}$
$\frac{v_{α}}{v_{P}} = \frac{1}{\sqrt{2}}$
$Or, v_{α} : v_{p} = 1 : \sqrt{2}$

J and K CET-2019