142473
If $E_{1}, E_{2}, E_{3}$ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 $E_{1}>E_{3}>E_{2}$
2 $E_{2}>E_{3}>E_{1}$
3 $\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
Explanation:
A Kinetic energy, $\mathrm{E}=\frac{1}{2} \mathrm{mv} v^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ According to the de- Broglie wavelength- $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Given, $\lambda_{1}=\lambda_{2}=\lambda_{3}$ $\therefore \quad \mathrm{E} \propto \frac{1}{\mathrm{~m}}$ $\because \quad \mathrm{m}_{\text {alpha }}>\mathrm{m}_{\text {proton }}>\mathrm{m}_{\text {electron }}$ So, lesser is the mass higher is the kinetic energy. $\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
BITSAT-2008
Dual nature of radiation and Matter
142474
The wavelength of $K_{\alpha}$-line characteristic $X$ rays emitted by an element is $0.32 \AA$. The wavelength of $K_{\beta}$-line emitted by the same element will be
1 $0.32 \AA$
2 $0.39 \AA$
3 $0.49 \AA$
4 $0.27 \AA$
Explanation:
D Given that, $\mathrm{k}_{\alpha}=0.32 \AA, \mathrm{k}_{\beta}=$ ? Energy of excitation is given by $\frac{1}{\lambda}=\mathrm{c}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $2 \rightarrow 1$ $\mathrm{k}_{\alpha}, \frac{1}{0.32}=\mathrm{c}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{c}}{4}$ $3 \rightarrow 1$ $\mathrm{k}_{\beta}, \frac{1}{\lambda}=\mathrm{c}\left(1-\frac{1}{9}\right)=\frac{8 \mathrm{c}}{9}$ Dividing equation (i) by (ii) we get- $\frac{\lambda}{0.32}=\frac{3 \mathrm{c}}{4} \times \frac{9}{8 \mathrm{c}}=\frac{27}{32}$ $\lambda=\frac{27}{32} \times 0.32=0.27 \AA$
BITSAT-2007
Dual nature of radiation and Matter
142476
If the momentum of electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be
1 $200 \mathrm{P}$
2 $400 \mathrm{P}$
3 $\frac{\mathrm{P}}{200}$
4 $100 \mathrm{P}$
Explanation:
A Given, change in de-Broglie wavelength $=$ $0.5 \%$ De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ Now, we differentiate the de-Broglie equation, $\mathrm{d} \lambda=-\frac{-\mathrm{h}}{\mathrm{P}^{2}} \mathrm{dP}$ But, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ $\mathrm{d} \lambda=\frac{-\lambda}{\mathrm{P}} \mathrm{dP}$ $\frac{\Delta \lambda}{\lambda}=\frac{\Delta \mathrm{P}}{\mathrm{p}}$ $\frac{0.5}{100}=\frac{\mathrm{P}}{\mathrm{P}_{\text {initial }}}$ $\mathrm{P}_{\text {initial }}=200 \mathrm{P}$
BITSAT-2005
Dual nature of radiation and Matter
142477
A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$. Then the wavelength of the de Broglie wave associated with it is:
B From de- Broglie wavelength, $\lambda=\frac{h}{p}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$ Here, $\quad \mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Since, $\quad \mathrm{v}=\mathrm{c}$ $\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{~m}=\frac{\mathrm{m}_{0}}{\sqrt{1-1}}=\frac{\mathrm{m}_{0}}{0}$ $\mathrm{m}=\infty$ Hence, $\lambda=\frac{\mathrm{h}}{\infty}=0$
BITSAT-2010
Dual nature of radiation and Matter
142478
An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is
1 $\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
2 $\frac{- \text { eht }}{\mathrm{E}}$
3 $\frac{-\mathrm{mh}}{\mathrm{eEt}^{2}}$
4 $\frac{-\mathrm{h}}{\mathrm{eE}}$
Explanation:
A We know that, The acceleration produced due to electric field is, $\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}$ Now, $v=u+$ at $\mathrm{v}=\mathrm{u}+\frac{\mathrm{eEt}}{\mathrm{m}} {[\because \mathrm{u}=0]}$ $\mathrm{v}=\frac{\mathrm{eEt}}{\mathrm{m}} \ldots \text { (ii) }$ de-Broglie wavelength of an electron is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $v$ from equation (ii) in equation (iii) $\lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{eEt}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{eEt}}$ Differentiating equation (iv) with respect to ' $t$ ' on both side $\therefore \quad \frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{\mathrm{h}}{-\mathrm{eEt}^{2}}$ $\frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
142473
If $E_{1}, E_{2}, E_{3}$ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 $E_{1}>E_{3}>E_{2}$
2 $E_{2}>E_{3}>E_{1}$
3 $\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
Explanation:
A Kinetic energy, $\mathrm{E}=\frac{1}{2} \mathrm{mv} v^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ According to the de- Broglie wavelength- $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Given, $\lambda_{1}=\lambda_{2}=\lambda_{3}$ $\therefore \quad \mathrm{E} \propto \frac{1}{\mathrm{~m}}$ $\because \quad \mathrm{m}_{\text {alpha }}>\mathrm{m}_{\text {proton }}>\mathrm{m}_{\text {electron }}$ So, lesser is the mass higher is the kinetic energy. $\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
BITSAT-2008
Dual nature of radiation and Matter
142474
The wavelength of $K_{\alpha}$-line characteristic $X$ rays emitted by an element is $0.32 \AA$. The wavelength of $K_{\beta}$-line emitted by the same element will be
1 $0.32 \AA$
2 $0.39 \AA$
3 $0.49 \AA$
4 $0.27 \AA$
Explanation:
D Given that, $\mathrm{k}_{\alpha}=0.32 \AA, \mathrm{k}_{\beta}=$ ? Energy of excitation is given by $\frac{1}{\lambda}=\mathrm{c}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $2 \rightarrow 1$ $\mathrm{k}_{\alpha}, \frac{1}{0.32}=\mathrm{c}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{c}}{4}$ $3 \rightarrow 1$ $\mathrm{k}_{\beta}, \frac{1}{\lambda}=\mathrm{c}\left(1-\frac{1}{9}\right)=\frac{8 \mathrm{c}}{9}$ Dividing equation (i) by (ii) we get- $\frac{\lambda}{0.32}=\frac{3 \mathrm{c}}{4} \times \frac{9}{8 \mathrm{c}}=\frac{27}{32}$ $\lambda=\frac{27}{32} \times 0.32=0.27 \AA$
BITSAT-2007
Dual nature of radiation and Matter
142476
If the momentum of electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be
1 $200 \mathrm{P}$
2 $400 \mathrm{P}$
3 $\frac{\mathrm{P}}{200}$
4 $100 \mathrm{P}$
Explanation:
A Given, change in de-Broglie wavelength $=$ $0.5 \%$ De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ Now, we differentiate the de-Broglie equation, $\mathrm{d} \lambda=-\frac{-\mathrm{h}}{\mathrm{P}^{2}} \mathrm{dP}$ But, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ $\mathrm{d} \lambda=\frac{-\lambda}{\mathrm{P}} \mathrm{dP}$ $\frac{\Delta \lambda}{\lambda}=\frac{\Delta \mathrm{P}}{\mathrm{p}}$ $\frac{0.5}{100}=\frac{\mathrm{P}}{\mathrm{P}_{\text {initial }}}$ $\mathrm{P}_{\text {initial }}=200 \mathrm{P}$
BITSAT-2005
Dual nature of radiation and Matter
142477
A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$. Then the wavelength of the de Broglie wave associated with it is:
B From de- Broglie wavelength, $\lambda=\frac{h}{p}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$ Here, $\quad \mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Since, $\quad \mathrm{v}=\mathrm{c}$ $\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{~m}=\frac{\mathrm{m}_{0}}{\sqrt{1-1}}=\frac{\mathrm{m}_{0}}{0}$ $\mathrm{m}=\infty$ Hence, $\lambda=\frac{\mathrm{h}}{\infty}=0$
BITSAT-2010
Dual nature of radiation and Matter
142478
An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is
1 $\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
2 $\frac{- \text { eht }}{\mathrm{E}}$
3 $\frac{-\mathrm{mh}}{\mathrm{eEt}^{2}}$
4 $\frac{-\mathrm{h}}{\mathrm{eE}}$
Explanation:
A We know that, The acceleration produced due to electric field is, $\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}$ Now, $v=u+$ at $\mathrm{v}=\mathrm{u}+\frac{\mathrm{eEt}}{\mathrm{m}} {[\because \mathrm{u}=0]}$ $\mathrm{v}=\frac{\mathrm{eEt}}{\mathrm{m}} \ldots \text { (ii) }$ de-Broglie wavelength of an electron is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $v$ from equation (ii) in equation (iii) $\lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{eEt}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{eEt}}$ Differentiating equation (iv) with respect to ' $t$ ' on both side $\therefore \quad \frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{\mathrm{h}}{-\mathrm{eEt}^{2}}$ $\frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
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Dual nature of radiation and Matter
142473
If $E_{1}, E_{2}, E_{3}$ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 $E_{1}>E_{3}>E_{2}$
2 $E_{2}>E_{3}>E_{1}$
3 $\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
Explanation:
A Kinetic energy, $\mathrm{E}=\frac{1}{2} \mathrm{mv} v^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ According to the de- Broglie wavelength- $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Given, $\lambda_{1}=\lambda_{2}=\lambda_{3}$ $\therefore \quad \mathrm{E} \propto \frac{1}{\mathrm{~m}}$ $\because \quad \mathrm{m}_{\text {alpha }}>\mathrm{m}_{\text {proton }}>\mathrm{m}_{\text {electron }}$ So, lesser is the mass higher is the kinetic energy. $\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
BITSAT-2008
Dual nature of radiation and Matter
142474
The wavelength of $K_{\alpha}$-line characteristic $X$ rays emitted by an element is $0.32 \AA$. The wavelength of $K_{\beta}$-line emitted by the same element will be
1 $0.32 \AA$
2 $0.39 \AA$
3 $0.49 \AA$
4 $0.27 \AA$
Explanation:
D Given that, $\mathrm{k}_{\alpha}=0.32 \AA, \mathrm{k}_{\beta}=$ ? Energy of excitation is given by $\frac{1}{\lambda}=\mathrm{c}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $2 \rightarrow 1$ $\mathrm{k}_{\alpha}, \frac{1}{0.32}=\mathrm{c}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{c}}{4}$ $3 \rightarrow 1$ $\mathrm{k}_{\beta}, \frac{1}{\lambda}=\mathrm{c}\left(1-\frac{1}{9}\right)=\frac{8 \mathrm{c}}{9}$ Dividing equation (i) by (ii) we get- $\frac{\lambda}{0.32}=\frac{3 \mathrm{c}}{4} \times \frac{9}{8 \mathrm{c}}=\frac{27}{32}$ $\lambda=\frac{27}{32} \times 0.32=0.27 \AA$
BITSAT-2007
Dual nature of radiation and Matter
142476
If the momentum of electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be
1 $200 \mathrm{P}$
2 $400 \mathrm{P}$
3 $\frac{\mathrm{P}}{200}$
4 $100 \mathrm{P}$
Explanation:
A Given, change in de-Broglie wavelength $=$ $0.5 \%$ De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ Now, we differentiate the de-Broglie equation, $\mathrm{d} \lambda=-\frac{-\mathrm{h}}{\mathrm{P}^{2}} \mathrm{dP}$ But, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ $\mathrm{d} \lambda=\frac{-\lambda}{\mathrm{P}} \mathrm{dP}$ $\frac{\Delta \lambda}{\lambda}=\frac{\Delta \mathrm{P}}{\mathrm{p}}$ $\frac{0.5}{100}=\frac{\mathrm{P}}{\mathrm{P}_{\text {initial }}}$ $\mathrm{P}_{\text {initial }}=200 \mathrm{P}$
BITSAT-2005
Dual nature of radiation and Matter
142477
A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$. Then the wavelength of the de Broglie wave associated with it is:
B From de- Broglie wavelength, $\lambda=\frac{h}{p}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$ Here, $\quad \mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Since, $\quad \mathrm{v}=\mathrm{c}$ $\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{~m}=\frac{\mathrm{m}_{0}}{\sqrt{1-1}}=\frac{\mathrm{m}_{0}}{0}$ $\mathrm{m}=\infty$ Hence, $\lambda=\frac{\mathrm{h}}{\infty}=0$
BITSAT-2010
Dual nature of radiation and Matter
142478
An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is
1 $\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
2 $\frac{- \text { eht }}{\mathrm{E}}$
3 $\frac{-\mathrm{mh}}{\mathrm{eEt}^{2}}$
4 $\frac{-\mathrm{h}}{\mathrm{eE}}$
Explanation:
A We know that, The acceleration produced due to electric field is, $\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}$ Now, $v=u+$ at $\mathrm{v}=\mathrm{u}+\frac{\mathrm{eEt}}{\mathrm{m}} {[\because \mathrm{u}=0]}$ $\mathrm{v}=\frac{\mathrm{eEt}}{\mathrm{m}} \ldots \text { (ii) }$ de-Broglie wavelength of an electron is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $v$ from equation (ii) in equation (iii) $\lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{eEt}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{eEt}}$ Differentiating equation (iv) with respect to ' $t$ ' on both side $\therefore \quad \frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{\mathrm{h}}{-\mathrm{eEt}^{2}}$ $\frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
142473
If $E_{1}, E_{2}, E_{3}$ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 $E_{1}>E_{3}>E_{2}$
2 $E_{2}>E_{3}>E_{1}$
3 $\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
Explanation:
A Kinetic energy, $\mathrm{E}=\frac{1}{2} \mathrm{mv} v^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ According to the de- Broglie wavelength- $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Given, $\lambda_{1}=\lambda_{2}=\lambda_{3}$ $\therefore \quad \mathrm{E} \propto \frac{1}{\mathrm{~m}}$ $\because \quad \mathrm{m}_{\text {alpha }}>\mathrm{m}_{\text {proton }}>\mathrm{m}_{\text {electron }}$ So, lesser is the mass higher is the kinetic energy. $\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
BITSAT-2008
Dual nature of radiation and Matter
142474
The wavelength of $K_{\alpha}$-line characteristic $X$ rays emitted by an element is $0.32 \AA$. The wavelength of $K_{\beta}$-line emitted by the same element will be
1 $0.32 \AA$
2 $0.39 \AA$
3 $0.49 \AA$
4 $0.27 \AA$
Explanation:
D Given that, $\mathrm{k}_{\alpha}=0.32 \AA, \mathrm{k}_{\beta}=$ ? Energy of excitation is given by $\frac{1}{\lambda}=\mathrm{c}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $2 \rightarrow 1$ $\mathrm{k}_{\alpha}, \frac{1}{0.32}=\mathrm{c}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{c}}{4}$ $3 \rightarrow 1$ $\mathrm{k}_{\beta}, \frac{1}{\lambda}=\mathrm{c}\left(1-\frac{1}{9}\right)=\frac{8 \mathrm{c}}{9}$ Dividing equation (i) by (ii) we get- $\frac{\lambda}{0.32}=\frac{3 \mathrm{c}}{4} \times \frac{9}{8 \mathrm{c}}=\frac{27}{32}$ $\lambda=\frac{27}{32} \times 0.32=0.27 \AA$
BITSAT-2007
Dual nature of radiation and Matter
142476
If the momentum of electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be
1 $200 \mathrm{P}$
2 $400 \mathrm{P}$
3 $\frac{\mathrm{P}}{200}$
4 $100 \mathrm{P}$
Explanation:
A Given, change in de-Broglie wavelength $=$ $0.5 \%$ De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ Now, we differentiate the de-Broglie equation, $\mathrm{d} \lambda=-\frac{-\mathrm{h}}{\mathrm{P}^{2}} \mathrm{dP}$ But, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ $\mathrm{d} \lambda=\frac{-\lambda}{\mathrm{P}} \mathrm{dP}$ $\frac{\Delta \lambda}{\lambda}=\frac{\Delta \mathrm{P}}{\mathrm{p}}$ $\frac{0.5}{100}=\frac{\mathrm{P}}{\mathrm{P}_{\text {initial }}}$ $\mathrm{P}_{\text {initial }}=200 \mathrm{P}$
BITSAT-2005
Dual nature of radiation and Matter
142477
A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$. Then the wavelength of the de Broglie wave associated with it is:
B From de- Broglie wavelength, $\lambda=\frac{h}{p}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$ Here, $\quad \mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Since, $\quad \mathrm{v}=\mathrm{c}$ $\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{~m}=\frac{\mathrm{m}_{0}}{\sqrt{1-1}}=\frac{\mathrm{m}_{0}}{0}$ $\mathrm{m}=\infty$ Hence, $\lambda=\frac{\mathrm{h}}{\infty}=0$
BITSAT-2010
Dual nature of radiation and Matter
142478
An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is
1 $\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
2 $\frac{- \text { eht }}{\mathrm{E}}$
3 $\frac{-\mathrm{mh}}{\mathrm{eEt}^{2}}$
4 $\frac{-\mathrm{h}}{\mathrm{eE}}$
Explanation:
A We know that, The acceleration produced due to electric field is, $\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}$ Now, $v=u+$ at $\mathrm{v}=\mathrm{u}+\frac{\mathrm{eEt}}{\mathrm{m}} {[\because \mathrm{u}=0]}$ $\mathrm{v}=\frac{\mathrm{eEt}}{\mathrm{m}} \ldots \text { (ii) }$ de-Broglie wavelength of an electron is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $v$ from equation (ii) in equation (iii) $\lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{eEt}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{eEt}}$ Differentiating equation (iv) with respect to ' $t$ ' on both side $\therefore \quad \frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{\mathrm{h}}{-\mathrm{eEt}^{2}}$ $\frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
142473
If $E_{1}, E_{2}, E_{3}$ are the respective kinetic energies of an electron, an alpha-particle and a proton, each having the same de-Broglie wavelength, then
1 $E_{1}>E_{3}>E_{2}$
2 $E_{2}>E_{3}>E_{1}$
3 $\mathrm{E}_{1}>\mathrm{E}_{2}>\mathrm{E}_{3}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}=\mathrm{E}_{3}$
Explanation:
A Kinetic energy, $\mathrm{E}=\frac{1}{2} \mathrm{mv} v^{2}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}$ According to the de- Broglie wavelength- $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Given, $\lambda_{1}=\lambda_{2}=\lambda_{3}$ $\therefore \quad \mathrm{E} \propto \frac{1}{\mathrm{~m}}$ $\because \quad \mathrm{m}_{\text {alpha }}>\mathrm{m}_{\text {proton }}>\mathrm{m}_{\text {electron }}$ So, lesser is the mass higher is the kinetic energy. $\mathrm{E}_{1}>\mathrm{E}_{3}>\mathrm{E}_{2}$
BITSAT-2008
Dual nature of radiation and Matter
142474
The wavelength of $K_{\alpha}$-line characteristic $X$ rays emitted by an element is $0.32 \AA$. The wavelength of $K_{\beta}$-line emitted by the same element will be
1 $0.32 \AA$
2 $0.39 \AA$
3 $0.49 \AA$
4 $0.27 \AA$
Explanation:
D Given that, $\mathrm{k}_{\alpha}=0.32 \AA, \mathrm{k}_{\beta}=$ ? Energy of excitation is given by $\frac{1}{\lambda}=\mathrm{c}\left(\frac{1}{\mathrm{n}_{2}^{2}}-\frac{1}{\mathrm{n}_{1}^{2}}\right)$ For $2 \rightarrow 1$ $\mathrm{k}_{\alpha}, \frac{1}{0.32}=\mathrm{c}\left(1-\frac{1}{4}\right)=\frac{3 \mathrm{c}}{4}$ $3 \rightarrow 1$ $\mathrm{k}_{\beta}, \frac{1}{\lambda}=\mathrm{c}\left(1-\frac{1}{9}\right)=\frac{8 \mathrm{c}}{9}$ Dividing equation (i) by (ii) we get- $\frac{\lambda}{0.32}=\frac{3 \mathrm{c}}{4} \times \frac{9}{8 \mathrm{c}}=\frac{27}{32}$ $\lambda=\frac{27}{32} \times 0.32=0.27 \AA$
BITSAT-2007
Dual nature of radiation and Matter
142476
If the momentum of electron is changed by $P$, then the de-Broglie wavelength associated with it changes by $0.5 \%$. The initial momentum of electron will be
1 $200 \mathrm{P}$
2 $400 \mathrm{P}$
3 $\frac{\mathrm{P}}{200}$
4 $100 \mathrm{P}$
Explanation:
A Given, change in de-Broglie wavelength $=$ $0.5 \%$ De-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ Now, we differentiate the de-Broglie equation, $\mathrm{d} \lambda=-\frac{-\mathrm{h}}{\mathrm{P}^{2}} \mathrm{dP}$ But, $\lambda=\frac{\mathrm{h}}{\mathrm{P}}$ $\mathrm{d} \lambda=\frac{-\lambda}{\mathrm{P}} \mathrm{dP}$ $\frac{\Delta \lambda}{\lambda}=\frac{\Delta \mathrm{P}}{\mathrm{p}}$ $\frac{0.5}{100}=\frac{\mathrm{P}}{\mathrm{P}_{\text {initial }}}$ $\mathrm{P}_{\text {initial }}=200 \mathrm{P}$
BITSAT-2005
Dual nature of radiation and Matter
142477
A material particle with a rest mass $m_{0}$ is moving with a velocity of light $c$. Then the wavelength of the de Broglie wave associated with it is:
B From de- Broglie wavelength, $\lambda=\frac{h}{p}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$ Here, $\quad \mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{v}^{2}}{\mathrm{c}^{2}}}}$ Since, $\quad \mathrm{v}=\mathrm{c}$ $\mathrm{m}=\frac{\mathrm{m}_{0}}{\sqrt{1-\frac{\mathrm{c}^{2}}{\mathrm{c}^{2}}}}$ $\mathrm{~m}=\frac{\mathrm{m}_{0}}{\sqrt{1-1}}=\frac{\mathrm{m}_{0}}{0}$ $\mathrm{m}=\infty$ Hence, $\lambda=\frac{\mathrm{h}}{\infty}=0$
BITSAT-2010
Dual nature of radiation and Matter
142478
An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field E. The rate of change of de-Broglie wavelength of this electron at time $t$ ignoring relativistic effects is
1 $\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$
2 $\frac{- \text { eht }}{\mathrm{E}}$
3 $\frac{-\mathrm{mh}}{\mathrm{eEt}^{2}}$
4 $\frac{-\mathrm{h}}{\mathrm{eE}}$
Explanation:
A We know that, The acceleration produced due to electric field is, $\mathrm{a}=\frac{\mathrm{eE}}{\mathrm{m}}$ Now, $v=u+$ at $\mathrm{v}=\mathrm{u}+\frac{\mathrm{eEt}}{\mathrm{m}} {[\because \mathrm{u}=0]}$ $\mathrm{v}=\frac{\mathrm{eEt}}{\mathrm{m}} \ldots \text { (ii) }$ de-Broglie wavelength of an electron is given by $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $v$ from equation (ii) in equation (iii) $\lambda=\frac{\mathrm{h}}{\mathrm{m} \frac{\mathrm{eEt}}{\mathrm{m}}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{eEt}}$ Differentiating equation (iv) with respect to ' $t$ ' on both side $\therefore \quad \frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{\mathrm{h}}{-\mathrm{eEt}^{2}}$ $\frac{\mathrm{d} \lambda}{\mathrm{dt}} =\frac{-\mathrm{h}}{\mathrm{eEt}^{2}}$