357866
The kinetic energy of an electron gets tripled then the de Broglie wavelength associated with it changes by a factor
1 \(\dfrac{1}{3}\)
2 \(\sqrt{3}\)
3 \(\dfrac{1}{\sqrt{3}}\)
4 3
Explanation:
For an electron, de Broglie wavelength \(\lambda = \frac{h}{{\sqrt {2mK} }}\) where \(h = \) Planck's constant, \(m = \) mass of an electron, \(k = \) kinetic energy of an electron Since \(h\) and \(m\) are constants, \(\lambda \propto \frac{1}{{\sqrt K }}\) \( \Rightarrow \frac{\lambda }{{{\lambda ^\prime }}} = \sqrt {\frac{{{K^\prime }}}{K}} = \sqrt {\frac{{3\;K}}{k}} {\rm{ or }}{\lambda ^\prime } = \frac{\lambda }{{\sqrt 3 }}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357867
de-Broglie wavelength of an electron accelerated by voltage of \(50\;V\) is close to \(\left( {|e| = 1.6 \times {{10}^{ - 19}}C,{m_e} = 9.1 \times {{10}^{ - 31}}\;kg} \right.\), \(\left. {h = 6.6 \times {{10}^{ - 34}}Js} \right)\):
357868
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
1 25
2 50
3 75
4 60
Explanation:
For de Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({1^{st}}\) case, \(\quad \lambda_{1}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({2^{nd}}\) case, \(\lambda_{2}=\dfrac{h}{\sqrt{2 m 16 K}}=\dfrac{h}{4 \sqrt{2 m K}}=\dfrac{\lambda_{1}}{4}\) So, \(75 \%\) change in the wavelength takes place.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357869
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is:
1 \(25 \%\)
2 \(75 \%\)
3 \(60 \%\)
4 \(50 \%\)
Explanation:
For de-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}} \Rightarrow \lambda \alpha \dfrac{1}{\sqrt{K}}\) Hence, \(\lambda_{2}=\dfrac{\lambda_{1}}{4}\) \(\dfrac{\lambda_{2}-\lambda_{1}}{\lambda_{1}} \times 100=75 \%\) There is \(75 \%\) change in the wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357870
A proton and an \(\alpha\)-particle are accelerated from rest by \(2\;V\) and \(4\;V\) potentials respectively. The ratio of their de-Broglie wavelength is
1 \(16: 1\)
2 \(8: 1\)
3 \(2: 1\)
4 \(4: 1\)
Explanation:
If a charged particle of mass \(m\), carrying a charge \(Q\) is accelerated through a potential difference of ' \(V\) ' volt, then the kinetic energy acquired by the particle is given by \(\dfrac{1}{2} m u^{2}=Q \cdot V \Rightarrow u=\sqrt{\dfrac{2 Q V}{m}}\) \(\therefore\) de Broglie wavelength of the charged particle is \(\lambda=\dfrac{h}{m V}=\dfrac{h}{\sqrt{2 Q m V} \text {. }}\) \(\therefore \dfrac{\lambda_{p}}{\lambda_{\alpha}} \sqrt{\dfrac{Q_{\alpha}}{Q_{p}}} \cdot \sqrt{\dfrac{m_{\alpha}}{m_{p}}} \cdot \sqrt{\dfrac{V_{\alpha}}{V_{p}}}\) \(=\sqrt{\dfrac{2 e}{e}} \cdot \sqrt{\dfrac{4 m p}{m p}} \cdot \sqrt{\dfrac{4 V}{2 V}}=4\).
357866
The kinetic energy of an electron gets tripled then the de Broglie wavelength associated with it changes by a factor
1 \(\dfrac{1}{3}\)
2 \(\sqrt{3}\)
3 \(\dfrac{1}{\sqrt{3}}\)
4 3
Explanation:
For an electron, de Broglie wavelength \(\lambda = \frac{h}{{\sqrt {2mK} }}\) where \(h = \) Planck's constant, \(m = \) mass of an electron, \(k = \) kinetic energy of an electron Since \(h\) and \(m\) are constants, \(\lambda \propto \frac{1}{{\sqrt K }}\) \( \Rightarrow \frac{\lambda }{{{\lambda ^\prime }}} = \sqrt {\frac{{{K^\prime }}}{K}} = \sqrt {\frac{{3\;K}}{k}} {\rm{ or }}{\lambda ^\prime } = \frac{\lambda }{{\sqrt 3 }}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357867
de-Broglie wavelength of an electron accelerated by voltage of \(50\;V\) is close to \(\left( {|e| = 1.6 \times {{10}^{ - 19}}C,{m_e} = 9.1 \times {{10}^{ - 31}}\;kg} \right.\), \(\left. {h = 6.6 \times {{10}^{ - 34}}Js} \right)\):
357868
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
1 25
2 50
3 75
4 60
Explanation:
For de Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({1^{st}}\) case, \(\quad \lambda_{1}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({2^{nd}}\) case, \(\lambda_{2}=\dfrac{h}{\sqrt{2 m 16 K}}=\dfrac{h}{4 \sqrt{2 m K}}=\dfrac{\lambda_{1}}{4}\) So, \(75 \%\) change in the wavelength takes place.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357869
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is:
1 \(25 \%\)
2 \(75 \%\)
3 \(60 \%\)
4 \(50 \%\)
Explanation:
For de-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}} \Rightarrow \lambda \alpha \dfrac{1}{\sqrt{K}}\) Hence, \(\lambda_{2}=\dfrac{\lambda_{1}}{4}\) \(\dfrac{\lambda_{2}-\lambda_{1}}{\lambda_{1}} \times 100=75 \%\) There is \(75 \%\) change in the wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357870
A proton and an \(\alpha\)-particle are accelerated from rest by \(2\;V\) and \(4\;V\) potentials respectively. The ratio of their de-Broglie wavelength is
1 \(16: 1\)
2 \(8: 1\)
3 \(2: 1\)
4 \(4: 1\)
Explanation:
If a charged particle of mass \(m\), carrying a charge \(Q\) is accelerated through a potential difference of ' \(V\) ' volt, then the kinetic energy acquired by the particle is given by \(\dfrac{1}{2} m u^{2}=Q \cdot V \Rightarrow u=\sqrt{\dfrac{2 Q V}{m}}\) \(\therefore\) de Broglie wavelength of the charged particle is \(\lambda=\dfrac{h}{m V}=\dfrac{h}{\sqrt{2 Q m V} \text {. }}\) \(\therefore \dfrac{\lambda_{p}}{\lambda_{\alpha}} \sqrt{\dfrac{Q_{\alpha}}{Q_{p}}} \cdot \sqrt{\dfrac{m_{\alpha}}{m_{p}}} \cdot \sqrt{\dfrac{V_{\alpha}}{V_{p}}}\) \(=\sqrt{\dfrac{2 e}{e}} \cdot \sqrt{\dfrac{4 m p}{m p}} \cdot \sqrt{\dfrac{4 V}{2 V}}=4\).
357866
The kinetic energy of an electron gets tripled then the de Broglie wavelength associated with it changes by a factor
1 \(\dfrac{1}{3}\)
2 \(\sqrt{3}\)
3 \(\dfrac{1}{\sqrt{3}}\)
4 3
Explanation:
For an electron, de Broglie wavelength \(\lambda = \frac{h}{{\sqrt {2mK} }}\) where \(h = \) Planck's constant, \(m = \) mass of an electron, \(k = \) kinetic energy of an electron Since \(h\) and \(m\) are constants, \(\lambda \propto \frac{1}{{\sqrt K }}\) \( \Rightarrow \frac{\lambda }{{{\lambda ^\prime }}} = \sqrt {\frac{{{K^\prime }}}{K}} = \sqrt {\frac{{3\;K}}{k}} {\rm{ or }}{\lambda ^\prime } = \frac{\lambda }{{\sqrt 3 }}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357867
de-Broglie wavelength of an electron accelerated by voltage of \(50\;V\) is close to \(\left( {|e| = 1.6 \times {{10}^{ - 19}}C,{m_e} = 9.1 \times {{10}^{ - 31}}\;kg} \right.\), \(\left. {h = 6.6 \times {{10}^{ - 34}}Js} \right)\):
357868
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
1 25
2 50
3 75
4 60
Explanation:
For de Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({1^{st}}\) case, \(\quad \lambda_{1}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({2^{nd}}\) case, \(\lambda_{2}=\dfrac{h}{\sqrt{2 m 16 K}}=\dfrac{h}{4 \sqrt{2 m K}}=\dfrac{\lambda_{1}}{4}\) So, \(75 \%\) change in the wavelength takes place.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357869
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is:
1 \(25 \%\)
2 \(75 \%\)
3 \(60 \%\)
4 \(50 \%\)
Explanation:
For de-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}} \Rightarrow \lambda \alpha \dfrac{1}{\sqrt{K}}\) Hence, \(\lambda_{2}=\dfrac{\lambda_{1}}{4}\) \(\dfrac{\lambda_{2}-\lambda_{1}}{\lambda_{1}} \times 100=75 \%\) There is \(75 \%\) change in the wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357870
A proton and an \(\alpha\)-particle are accelerated from rest by \(2\;V\) and \(4\;V\) potentials respectively. The ratio of their de-Broglie wavelength is
1 \(16: 1\)
2 \(8: 1\)
3 \(2: 1\)
4 \(4: 1\)
Explanation:
If a charged particle of mass \(m\), carrying a charge \(Q\) is accelerated through a potential difference of ' \(V\) ' volt, then the kinetic energy acquired by the particle is given by \(\dfrac{1}{2} m u^{2}=Q \cdot V \Rightarrow u=\sqrt{\dfrac{2 Q V}{m}}\) \(\therefore\) de Broglie wavelength of the charged particle is \(\lambda=\dfrac{h}{m V}=\dfrac{h}{\sqrt{2 Q m V} \text {. }}\) \(\therefore \dfrac{\lambda_{p}}{\lambda_{\alpha}} \sqrt{\dfrac{Q_{\alpha}}{Q_{p}}} \cdot \sqrt{\dfrac{m_{\alpha}}{m_{p}}} \cdot \sqrt{\dfrac{V_{\alpha}}{V_{p}}}\) \(=\sqrt{\dfrac{2 e}{e}} \cdot \sqrt{\dfrac{4 m p}{m p}} \cdot \sqrt{\dfrac{4 V}{2 V}}=4\).
357866
The kinetic energy of an electron gets tripled then the de Broglie wavelength associated with it changes by a factor
1 \(\dfrac{1}{3}\)
2 \(\sqrt{3}\)
3 \(\dfrac{1}{\sqrt{3}}\)
4 3
Explanation:
For an electron, de Broglie wavelength \(\lambda = \frac{h}{{\sqrt {2mK} }}\) where \(h = \) Planck's constant, \(m = \) mass of an electron, \(k = \) kinetic energy of an electron Since \(h\) and \(m\) are constants, \(\lambda \propto \frac{1}{{\sqrt K }}\) \( \Rightarrow \frac{\lambda }{{{\lambda ^\prime }}} = \sqrt {\frac{{{K^\prime }}}{K}} = \sqrt {\frac{{3\;K}}{k}} {\rm{ or }}{\lambda ^\prime } = \frac{\lambda }{{\sqrt 3 }}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357867
de-Broglie wavelength of an electron accelerated by voltage of \(50\;V\) is close to \(\left( {|e| = 1.6 \times {{10}^{ - 19}}C,{m_e} = 9.1 \times {{10}^{ - 31}}\;kg} \right.\), \(\left. {h = 6.6 \times {{10}^{ - 34}}Js} \right)\):
357868
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
1 25
2 50
3 75
4 60
Explanation:
For de Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({1^{st}}\) case, \(\quad \lambda_{1}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({2^{nd}}\) case, \(\lambda_{2}=\dfrac{h}{\sqrt{2 m 16 K}}=\dfrac{h}{4 \sqrt{2 m K}}=\dfrac{\lambda_{1}}{4}\) So, \(75 \%\) change in the wavelength takes place.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357869
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is:
1 \(25 \%\)
2 \(75 \%\)
3 \(60 \%\)
4 \(50 \%\)
Explanation:
For de-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}} \Rightarrow \lambda \alpha \dfrac{1}{\sqrt{K}}\) Hence, \(\lambda_{2}=\dfrac{\lambda_{1}}{4}\) \(\dfrac{\lambda_{2}-\lambda_{1}}{\lambda_{1}} \times 100=75 \%\) There is \(75 \%\) change in the wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357870
A proton and an \(\alpha\)-particle are accelerated from rest by \(2\;V\) and \(4\;V\) potentials respectively. The ratio of their de-Broglie wavelength is
1 \(16: 1\)
2 \(8: 1\)
3 \(2: 1\)
4 \(4: 1\)
Explanation:
If a charged particle of mass \(m\), carrying a charge \(Q\) is accelerated through a potential difference of ' \(V\) ' volt, then the kinetic energy acquired by the particle is given by \(\dfrac{1}{2} m u^{2}=Q \cdot V \Rightarrow u=\sqrt{\dfrac{2 Q V}{m}}\) \(\therefore\) de Broglie wavelength of the charged particle is \(\lambda=\dfrac{h}{m V}=\dfrac{h}{\sqrt{2 Q m V} \text {. }}\) \(\therefore \dfrac{\lambda_{p}}{\lambda_{\alpha}} \sqrt{\dfrac{Q_{\alpha}}{Q_{p}}} \cdot \sqrt{\dfrac{m_{\alpha}}{m_{p}}} \cdot \sqrt{\dfrac{V_{\alpha}}{V_{p}}}\) \(=\sqrt{\dfrac{2 e}{e}} \cdot \sqrt{\dfrac{4 m p}{m p}} \cdot \sqrt{\dfrac{4 V}{2 V}}=4\).
357866
The kinetic energy of an electron gets tripled then the de Broglie wavelength associated with it changes by a factor
1 \(\dfrac{1}{3}\)
2 \(\sqrt{3}\)
3 \(\dfrac{1}{\sqrt{3}}\)
4 3
Explanation:
For an electron, de Broglie wavelength \(\lambda = \frac{h}{{\sqrt {2mK} }}\) where \(h = \) Planck's constant, \(m = \) mass of an electron, \(k = \) kinetic energy of an electron Since \(h\) and \(m\) are constants, \(\lambda \propto \frac{1}{{\sqrt K }}\) \( \Rightarrow \frac{\lambda }{{{\lambda ^\prime }}} = \sqrt {\frac{{{K^\prime }}}{K}} = \sqrt {\frac{{3\;K}}{k}} {\rm{ or }}{\lambda ^\prime } = \frac{\lambda }{{\sqrt 3 }}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357867
de-Broglie wavelength of an electron accelerated by voltage of \(50\;V\) is close to \(\left( {|e| = 1.6 \times {{10}^{ - 19}}C,{m_e} = 9.1 \times {{10}^{ - 31}}\;kg} \right.\), \(\left. {h = 6.6 \times {{10}^{ - 34}}Js} \right)\):
357868
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de Broglie wavelength of the particle is
1 25
2 50
3 75
4 60
Explanation:
For de Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({1^{st}}\) case, \(\quad \lambda_{1}=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}}\) For \({2^{nd}}\) case, \(\lambda_{2}=\dfrac{h}{\sqrt{2 m 16 K}}=\dfrac{h}{4 \sqrt{2 m K}}=\dfrac{\lambda_{1}}{4}\) So, \(75 \%\) change in the wavelength takes place.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357869
If the kinetic energy of the particle is increased by 16 times, the percentage change in the de-Broglie wavelength of the particle is:
1 \(25 \%\)
2 \(75 \%\)
3 \(60 \%\)
4 \(50 \%\)
Explanation:
For de-Broglie wavelength, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m K}} \Rightarrow \lambda \alpha \dfrac{1}{\sqrt{K}}\) Hence, \(\lambda_{2}=\dfrac{\lambda_{1}}{4}\) \(\dfrac{\lambda_{2}-\lambda_{1}}{\lambda_{1}} \times 100=75 \%\) There is \(75 \%\) change in the wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357870
A proton and an \(\alpha\)-particle are accelerated from rest by \(2\;V\) and \(4\;V\) potentials respectively. The ratio of their de-Broglie wavelength is
1 \(16: 1\)
2 \(8: 1\)
3 \(2: 1\)
4 \(4: 1\)
Explanation:
If a charged particle of mass \(m\), carrying a charge \(Q\) is accelerated through a potential difference of ' \(V\) ' volt, then the kinetic energy acquired by the particle is given by \(\dfrac{1}{2} m u^{2}=Q \cdot V \Rightarrow u=\sqrt{\dfrac{2 Q V}{m}}\) \(\therefore\) de Broglie wavelength of the charged particle is \(\lambda=\dfrac{h}{m V}=\dfrac{h}{\sqrt{2 Q m V} \text {. }}\) \(\therefore \dfrac{\lambda_{p}}{\lambda_{\alpha}} \sqrt{\dfrac{Q_{\alpha}}{Q_{p}}} \cdot \sqrt{\dfrac{m_{\alpha}}{m_{p}}} \cdot \sqrt{\dfrac{V_{\alpha}}{V_{p}}}\) \(=\sqrt{\dfrac{2 e}{e}} \cdot \sqrt{\dfrac{4 m p}{m p}} \cdot \sqrt{\dfrac{4 V}{2 V}}=4\).
