357836
The energy of an electron having de - Brogile wavelength \(\lambda\) is ( where \(h=\) Planck's constant, \(\mathrm{m}=\) mass of electron)
1 \(\dfrac{h}{2 m \lambda}\)
2 \(\dfrac{h^{2}}{2 m \lambda^{2}}\)
3 \(\dfrac{h^{2}}{2 m^{2} \lambda^{2}}\)
4 \(\dfrac{h^{2}}{2 m^{2} \lambda}\)
Explanation:
We know that \(\lambda=\dfrac{h}{\sqrt{2 m(K E)}}\) on taking square on both sides, we get \({\lambda ^2} = \frac{{{h^2}}}{{2m(KE)}}\) \(KE = \frac{{{h^2}}}{{2m{\lambda ^2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357837
A proton and an \(\alpha\) particle are accelerated through the same potential difference \(V\). The ratio of their de Broglie wavelength is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 \(\sqrt{3}\)
4 \(2 \sqrt{3}\)
Explanation:
de Broglie wavelength of a particle of mass \(m\) and charge \(q\) accelerated through a potential difference \(V\) is \(\lambda = \frac{h}{{\sqrt {2mqV} }}\) Since the potential difference \(V\) is the same for both the particles, \(\lambda \propto \frac{1}{{\sqrt {mq} }}\) Thus \(\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\left(\dfrac{4 m_{p}}{m_{p}}\right)\left(\dfrac{2 e}{e}\right)}=\dfrac{\sqrt{8}}{1}=2 \sqrt{2}\)
KCET - 2018
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357838
Calculate de-Broglie wavelength for an average helium atom in a furnace at \(400\,K.\) Given mass of helium \( = 4.002\,amu.\)
1 0.35
2 0.63
3 0.92
4 0.43
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\) Mean \(KE\) of helium atom \(=\dfrac{3 k T}{2}\) or \(E=\dfrac{3 \times\left(1.38 \times 10^{-23}\right) \times 400}{2}\) In terms of momentum, the energy is given by \(E=\dfrac{p^{2}}{2 m}\) or \(\quad p^{2}=2 m E\) or \(p=\sqrt{2 m E}\) \(\therefore \quad p = \sqrt {\begin{array}{*{20}{l}}{2 \times \left( {4.002 \times 1.66 \times {{10}^{ - 27}}} \right) \times 3 \times }\\{\left( {1.38 \times {{10}^{ - 23}}} \right) \times 200}\end{array}} \) (where mass of helium \(4.000\,amu = 4.002 \times 1.66 \times {10^{ - 27}}\;kg\,)\) Now, \(\lambda=\dfrac{6.625 \times 10^{-34}}{p}\) Substituting the value of \(p\) and solving it, we get \(\lambda=0.63 \times 10^{-10} {~m}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357839
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
1 More than that of a proton
2 Equal to that of a proton
3 Zero
4 Less than that of a proton.
Explanation:
Using \(E=\dfrac{p^{2}}{2 m}\) and \(\lambda=\dfrac{h}{p}\), \(\begin{gathered}\quad E=\dfrac{h^{2}}{2 m \lambda^{2}} \\\therefore \text { For same } \lambda, \dfrac{E_{e}}{E_{p}}=\dfrac{m_{p}}{m_{e}},\end{gathered}\) As \({m_p} > > {m_e},{E_e} > > {E_p}\).
KCET - 2008
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357840
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume \(m_{p}=m_{e} \times 1849\))
1 \(2: 43\)
2 \(1: 43\)
3 \(1: 30\)
4 \(1: 62\)
Explanation:
Kinetic energy, of a particle of mass \(m\) is given by \(E_{k}=\dfrac{1}{2} m v^{2}\) where \(v\) is the velocity of the particle. As, \(p=m v=\sqrt{2 m E_{k}}\) \(\therefore\) de-broglie wavelength of the moving particle is given by, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E_{k}}}\) For proton, \(\lambda_{p}=\dfrac{h}{\sqrt{2 m_{p} E_{k}}}\); for electron, \(\lambda_{e}=\dfrac{h}{\sqrt{2 m_{e} E_{k}}}\) So, \(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\sqrt{\dfrac{m_{e}}{m_{e} \times 1849}}\); \(\dfrac{\lambda_{p}}{\lambda_{e}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}\)
357836
The energy of an electron having de - Brogile wavelength \(\lambda\) is ( where \(h=\) Planck's constant, \(\mathrm{m}=\) mass of electron)
1 \(\dfrac{h}{2 m \lambda}\)
2 \(\dfrac{h^{2}}{2 m \lambda^{2}}\)
3 \(\dfrac{h^{2}}{2 m^{2} \lambda^{2}}\)
4 \(\dfrac{h^{2}}{2 m^{2} \lambda}\)
Explanation:
We know that \(\lambda=\dfrac{h}{\sqrt{2 m(K E)}}\) on taking square on both sides, we get \({\lambda ^2} = \frac{{{h^2}}}{{2m(KE)}}\) \(KE = \frac{{{h^2}}}{{2m{\lambda ^2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357837
A proton and an \(\alpha\) particle are accelerated through the same potential difference \(V\). The ratio of their de Broglie wavelength is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 \(\sqrt{3}\)
4 \(2 \sqrt{3}\)
Explanation:
de Broglie wavelength of a particle of mass \(m\) and charge \(q\) accelerated through a potential difference \(V\) is \(\lambda = \frac{h}{{\sqrt {2mqV} }}\) Since the potential difference \(V\) is the same for both the particles, \(\lambda \propto \frac{1}{{\sqrt {mq} }}\) Thus \(\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\left(\dfrac{4 m_{p}}{m_{p}}\right)\left(\dfrac{2 e}{e}\right)}=\dfrac{\sqrt{8}}{1}=2 \sqrt{2}\)
KCET - 2018
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357838
Calculate de-Broglie wavelength for an average helium atom in a furnace at \(400\,K.\) Given mass of helium \( = 4.002\,amu.\)
1 0.35
2 0.63
3 0.92
4 0.43
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\) Mean \(KE\) of helium atom \(=\dfrac{3 k T}{2}\) or \(E=\dfrac{3 \times\left(1.38 \times 10^{-23}\right) \times 400}{2}\) In terms of momentum, the energy is given by \(E=\dfrac{p^{2}}{2 m}\) or \(\quad p^{2}=2 m E\) or \(p=\sqrt{2 m E}\) \(\therefore \quad p = \sqrt {\begin{array}{*{20}{l}}{2 \times \left( {4.002 \times 1.66 \times {{10}^{ - 27}}} \right) \times 3 \times }\\{\left( {1.38 \times {{10}^{ - 23}}} \right) \times 200}\end{array}} \) (where mass of helium \(4.000\,amu = 4.002 \times 1.66 \times {10^{ - 27}}\;kg\,)\) Now, \(\lambda=\dfrac{6.625 \times 10^{-34}}{p}\) Substituting the value of \(p\) and solving it, we get \(\lambda=0.63 \times 10^{-10} {~m}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357839
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
1 More than that of a proton
2 Equal to that of a proton
3 Zero
4 Less than that of a proton.
Explanation:
Using \(E=\dfrac{p^{2}}{2 m}\) and \(\lambda=\dfrac{h}{p}\), \(\begin{gathered}\quad E=\dfrac{h^{2}}{2 m \lambda^{2}} \\\therefore \text { For same } \lambda, \dfrac{E_{e}}{E_{p}}=\dfrac{m_{p}}{m_{e}},\end{gathered}\) As \({m_p} > > {m_e},{E_e} > > {E_p}\).
KCET - 2008
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357840
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume \(m_{p}=m_{e} \times 1849\))
1 \(2: 43\)
2 \(1: 43\)
3 \(1: 30\)
4 \(1: 62\)
Explanation:
Kinetic energy, of a particle of mass \(m\) is given by \(E_{k}=\dfrac{1}{2} m v^{2}\) where \(v\) is the velocity of the particle. As, \(p=m v=\sqrt{2 m E_{k}}\) \(\therefore\) de-broglie wavelength of the moving particle is given by, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E_{k}}}\) For proton, \(\lambda_{p}=\dfrac{h}{\sqrt{2 m_{p} E_{k}}}\); for electron, \(\lambda_{e}=\dfrac{h}{\sqrt{2 m_{e} E_{k}}}\) So, \(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\sqrt{\dfrac{m_{e}}{m_{e} \times 1849}}\); \(\dfrac{\lambda_{p}}{\lambda_{e}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}\)
357836
The energy of an electron having de - Brogile wavelength \(\lambda\) is ( where \(h=\) Planck's constant, \(\mathrm{m}=\) mass of electron)
1 \(\dfrac{h}{2 m \lambda}\)
2 \(\dfrac{h^{2}}{2 m \lambda^{2}}\)
3 \(\dfrac{h^{2}}{2 m^{2} \lambda^{2}}\)
4 \(\dfrac{h^{2}}{2 m^{2} \lambda}\)
Explanation:
We know that \(\lambda=\dfrac{h}{\sqrt{2 m(K E)}}\) on taking square on both sides, we get \({\lambda ^2} = \frac{{{h^2}}}{{2m(KE)}}\) \(KE = \frac{{{h^2}}}{{2m{\lambda ^2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357837
A proton and an \(\alpha\) particle are accelerated through the same potential difference \(V\). The ratio of their de Broglie wavelength is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 \(\sqrt{3}\)
4 \(2 \sqrt{3}\)
Explanation:
de Broglie wavelength of a particle of mass \(m\) and charge \(q\) accelerated through a potential difference \(V\) is \(\lambda = \frac{h}{{\sqrt {2mqV} }}\) Since the potential difference \(V\) is the same for both the particles, \(\lambda \propto \frac{1}{{\sqrt {mq} }}\) Thus \(\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\left(\dfrac{4 m_{p}}{m_{p}}\right)\left(\dfrac{2 e}{e}\right)}=\dfrac{\sqrt{8}}{1}=2 \sqrt{2}\)
KCET - 2018
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357838
Calculate de-Broglie wavelength for an average helium atom in a furnace at \(400\,K.\) Given mass of helium \( = 4.002\,amu.\)
1 0.35
2 0.63
3 0.92
4 0.43
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\) Mean \(KE\) of helium atom \(=\dfrac{3 k T}{2}\) or \(E=\dfrac{3 \times\left(1.38 \times 10^{-23}\right) \times 400}{2}\) In terms of momentum, the energy is given by \(E=\dfrac{p^{2}}{2 m}\) or \(\quad p^{2}=2 m E\) or \(p=\sqrt{2 m E}\) \(\therefore \quad p = \sqrt {\begin{array}{*{20}{l}}{2 \times \left( {4.002 \times 1.66 \times {{10}^{ - 27}}} \right) \times 3 \times }\\{\left( {1.38 \times {{10}^{ - 23}}} \right) \times 200}\end{array}} \) (where mass of helium \(4.000\,amu = 4.002 \times 1.66 \times {10^{ - 27}}\;kg\,)\) Now, \(\lambda=\dfrac{6.625 \times 10^{-34}}{p}\) Substituting the value of \(p\) and solving it, we get \(\lambda=0.63 \times 10^{-10} {~m}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357839
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
1 More than that of a proton
2 Equal to that of a proton
3 Zero
4 Less than that of a proton.
Explanation:
Using \(E=\dfrac{p^{2}}{2 m}\) and \(\lambda=\dfrac{h}{p}\), \(\begin{gathered}\quad E=\dfrac{h^{2}}{2 m \lambda^{2}} \\\therefore \text { For same } \lambda, \dfrac{E_{e}}{E_{p}}=\dfrac{m_{p}}{m_{e}},\end{gathered}\) As \({m_p} > > {m_e},{E_e} > > {E_p}\).
KCET - 2008
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357840
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume \(m_{p}=m_{e} \times 1849\))
1 \(2: 43\)
2 \(1: 43\)
3 \(1: 30\)
4 \(1: 62\)
Explanation:
Kinetic energy, of a particle of mass \(m\) is given by \(E_{k}=\dfrac{1}{2} m v^{2}\) where \(v\) is the velocity of the particle. As, \(p=m v=\sqrt{2 m E_{k}}\) \(\therefore\) de-broglie wavelength of the moving particle is given by, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E_{k}}}\) For proton, \(\lambda_{p}=\dfrac{h}{\sqrt{2 m_{p} E_{k}}}\); for electron, \(\lambda_{e}=\dfrac{h}{\sqrt{2 m_{e} E_{k}}}\) So, \(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\sqrt{\dfrac{m_{e}}{m_{e} \times 1849}}\); \(\dfrac{\lambda_{p}}{\lambda_{e}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357836
The energy of an electron having de - Brogile wavelength \(\lambda\) is ( where \(h=\) Planck's constant, \(\mathrm{m}=\) mass of electron)
1 \(\dfrac{h}{2 m \lambda}\)
2 \(\dfrac{h^{2}}{2 m \lambda^{2}}\)
3 \(\dfrac{h^{2}}{2 m^{2} \lambda^{2}}\)
4 \(\dfrac{h^{2}}{2 m^{2} \lambda}\)
Explanation:
We know that \(\lambda=\dfrac{h}{\sqrt{2 m(K E)}}\) on taking square on both sides, we get \({\lambda ^2} = \frac{{{h^2}}}{{2m(KE)}}\) \(KE = \frac{{{h^2}}}{{2m{\lambda ^2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357837
A proton and an \(\alpha\) particle are accelerated through the same potential difference \(V\). The ratio of their de Broglie wavelength is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 \(\sqrt{3}\)
4 \(2 \sqrt{3}\)
Explanation:
de Broglie wavelength of a particle of mass \(m\) and charge \(q\) accelerated through a potential difference \(V\) is \(\lambda = \frac{h}{{\sqrt {2mqV} }}\) Since the potential difference \(V\) is the same for both the particles, \(\lambda \propto \frac{1}{{\sqrt {mq} }}\) Thus \(\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\left(\dfrac{4 m_{p}}{m_{p}}\right)\left(\dfrac{2 e}{e}\right)}=\dfrac{\sqrt{8}}{1}=2 \sqrt{2}\)
KCET - 2018
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357838
Calculate de-Broglie wavelength for an average helium atom in a furnace at \(400\,K.\) Given mass of helium \( = 4.002\,amu.\)
1 0.35
2 0.63
3 0.92
4 0.43
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\) Mean \(KE\) of helium atom \(=\dfrac{3 k T}{2}\) or \(E=\dfrac{3 \times\left(1.38 \times 10^{-23}\right) \times 400}{2}\) In terms of momentum, the energy is given by \(E=\dfrac{p^{2}}{2 m}\) or \(\quad p^{2}=2 m E\) or \(p=\sqrt{2 m E}\) \(\therefore \quad p = \sqrt {\begin{array}{*{20}{l}}{2 \times \left( {4.002 \times 1.66 \times {{10}^{ - 27}}} \right) \times 3 \times }\\{\left( {1.38 \times {{10}^{ - 23}}} \right) \times 200}\end{array}} \) (where mass of helium \(4.000\,amu = 4.002 \times 1.66 \times {10^{ - 27}}\;kg\,)\) Now, \(\lambda=\dfrac{6.625 \times 10^{-34}}{p}\) Substituting the value of \(p\) and solving it, we get \(\lambda=0.63 \times 10^{-10} {~m}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357839
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
1 More than that of a proton
2 Equal to that of a proton
3 Zero
4 Less than that of a proton.
Explanation:
Using \(E=\dfrac{p^{2}}{2 m}\) and \(\lambda=\dfrac{h}{p}\), \(\begin{gathered}\quad E=\dfrac{h^{2}}{2 m \lambda^{2}} \\\therefore \text { For same } \lambda, \dfrac{E_{e}}{E_{p}}=\dfrac{m_{p}}{m_{e}},\end{gathered}\) As \({m_p} > > {m_e},{E_e} > > {E_p}\).
KCET - 2008
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357840
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume \(m_{p}=m_{e} \times 1849\))
1 \(2: 43\)
2 \(1: 43\)
3 \(1: 30\)
4 \(1: 62\)
Explanation:
Kinetic energy, of a particle of mass \(m\) is given by \(E_{k}=\dfrac{1}{2} m v^{2}\) where \(v\) is the velocity of the particle. As, \(p=m v=\sqrt{2 m E_{k}}\) \(\therefore\) de-broglie wavelength of the moving particle is given by, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E_{k}}}\) For proton, \(\lambda_{p}=\dfrac{h}{\sqrt{2 m_{p} E_{k}}}\); for electron, \(\lambda_{e}=\dfrac{h}{\sqrt{2 m_{e} E_{k}}}\) So, \(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\sqrt{\dfrac{m_{e}}{m_{e} \times 1849}}\); \(\dfrac{\lambda_{p}}{\lambda_{e}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}\)
357836
The energy of an electron having de - Brogile wavelength \(\lambda\) is ( where \(h=\) Planck's constant, \(\mathrm{m}=\) mass of electron)
1 \(\dfrac{h}{2 m \lambda}\)
2 \(\dfrac{h^{2}}{2 m \lambda^{2}}\)
3 \(\dfrac{h^{2}}{2 m^{2} \lambda^{2}}\)
4 \(\dfrac{h^{2}}{2 m^{2} \lambda}\)
Explanation:
We know that \(\lambda=\dfrac{h}{\sqrt{2 m(K E)}}\) on taking square on both sides, we get \({\lambda ^2} = \frac{{{h^2}}}{{2m(KE)}}\) \(KE = \frac{{{h^2}}}{{2m{\lambda ^2}}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357837
A proton and an \(\alpha\) particle are accelerated through the same potential difference \(V\). The ratio of their de Broglie wavelength is
1 \(\sqrt{2}\)
2 \(2 \sqrt{2}\)
3 \(\sqrt{3}\)
4 \(2 \sqrt{3}\)
Explanation:
de Broglie wavelength of a particle of mass \(m\) and charge \(q\) accelerated through a potential difference \(V\) is \(\lambda = \frac{h}{{\sqrt {2mqV} }}\) Since the potential difference \(V\) is the same for both the particles, \(\lambda \propto \frac{1}{{\sqrt {mq} }}\) Thus \(\dfrac{\lambda_{p}}{\lambda_{\alpha}}=\sqrt{\dfrac{m_{\alpha} q_{\alpha}}{m_{p} q_{p}}}=\sqrt{\left(\dfrac{4 m_{p}}{m_{p}}\right)\left(\dfrac{2 e}{e}\right)}=\dfrac{\sqrt{8}}{1}=2 \sqrt{2}\)
KCET - 2018
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357838
Calculate de-Broglie wavelength for an average helium atom in a furnace at \(400\,K.\) Given mass of helium \( = 4.002\,amu.\)
1 0.35
2 0.63
3 0.92
4 0.43
Explanation:
de-Broglie wavelength, \(\lambda=\dfrac{h}{p}\) Mean \(KE\) of helium atom \(=\dfrac{3 k T}{2}\) or \(E=\dfrac{3 \times\left(1.38 \times 10^{-23}\right) \times 400}{2}\) In terms of momentum, the energy is given by \(E=\dfrac{p^{2}}{2 m}\) or \(\quad p^{2}=2 m E\) or \(p=\sqrt{2 m E}\) \(\therefore \quad p = \sqrt {\begin{array}{*{20}{l}}{2 \times \left( {4.002 \times 1.66 \times {{10}^{ - 27}}} \right) \times 3 \times }\\{\left( {1.38 \times {{10}^{ - 23}}} \right) \times 200}\end{array}} \) (where mass of helium \(4.000\,amu = 4.002 \times 1.66 \times {10^{ - 27}}\;kg\,)\) Now, \(\lambda=\dfrac{6.625 \times 10^{-34}}{p}\) Substituting the value of \(p\) and solving it, we get \(\lambda=0.63 \times 10^{-10} {~m}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357839
If an electron and a proton have the same de-Broglie wavelength, then the kinetic energy of the electron is
1 More than that of a proton
2 Equal to that of a proton
3 Zero
4 Less than that of a proton.
Explanation:
Using \(E=\dfrac{p^{2}}{2 m}\) and \(\lambda=\dfrac{h}{p}\), \(\begin{gathered}\quad E=\dfrac{h^{2}}{2 m \lambda^{2}} \\\therefore \text { For same } \lambda, \dfrac{E_{e}}{E_{p}}=\dfrac{m_{p}}{m_{e}},\end{gathered}\) As \({m_p} > > {m_e},{E_e} > > {E_p}\).
KCET - 2008
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357840
The ratio of the de-Broglie wavelengths of proton and electron having same kinetic energy (Assume \(m_{p}=m_{e} \times 1849\))
1 \(2: 43\)
2 \(1: 43\)
3 \(1: 30\)
4 \(1: 62\)
Explanation:
Kinetic energy, of a particle of mass \(m\) is given by \(E_{k}=\dfrac{1}{2} m v^{2}\) where \(v\) is the velocity of the particle. As, \(p=m v=\sqrt{2 m E_{k}}\) \(\therefore\) de-broglie wavelength of the moving particle is given by, \(\lambda=\dfrac{h}{p}=\dfrac{h}{\sqrt{2 m E_{k}}}\) For proton, \(\lambda_{p}=\dfrac{h}{\sqrt{2 m_{p} E_{k}}}\); for electron, \(\lambda_{e}=\dfrac{h}{\sqrt{2 m_{e} E_{k}}}\) So, \(\dfrac{\lambda_{p}}{\lambda_{e}}=\dfrac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\sqrt{\dfrac{m_{e}}{m_{e} \times 1849}}\); \(\dfrac{\lambda_{p}}{\lambda_{e}}=\sqrt{\dfrac{1}{1849}}=\dfrac{1}{43}\)