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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357909 The figure shows standing de Broglie waves due to the revolvtion of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their ususal meanings)
supporting img

1 \(\frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
2 \(\frac{{4\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
3 \(\frac{{9\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
4 \(\frac{{36\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
KCET - 2021
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357910 The de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to

1 radius of orbit
2 perimeter of orbit
3 diameter of orbit
4 half the perimeter of orbit
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357911 An electron (of mass \(m\) ) with an initial velocity \(v = {v_0}\widehat i\) is moving in an electric field \(E = {E_0}\widehat j\). If \({\lambda _0} = h/m{v_0}\), it's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0} \sqrt{1+\dfrac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\)
2 \(\lambda_{0}\)
3 \(\frac{{{\lambda _0}}}{{\left( {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right)}}\)
4 \(\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357912 The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4 K, 2 K\) and \(K\) respectively. The de-Broglie wavelength associated with electron \(\left(\lambda_{e}\right), \alpha\)-particle \(\left(\lambda_{\alpha}\right)\) and the proton \(\left(\lambda_{p}\right)\) are as follows :

1 \(\lambda_{\alpha}>\lambda_{p}>\lambda_{e}\)
2 \(\lambda_{\alpha}=\lambda_{p} < \lambda_{e}\)
3 \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\)
4 \(\lambda_{\alpha}=\lambda_{p}>\lambda_{e}\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357913 The ratio of momentum of an electron and \(\alpha\)-particle which are accelerated from rest by a potential difference of \(100\;V\) is

1 1
2 \(\sqrt{\left(\dfrac{2 m_{e}}{m_{\alpha}}\right)}\)
3 \(\sqrt{\left(\dfrac{m_{e}}{m_{\alpha}}\right)}\)
4 \(\sqrt{\left(\dfrac{m_{e}}{2 m_{\alpha}}\right)}\)
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357909 The figure shows standing de Broglie waves due to the revolvtion of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their ususal meanings)
supporting img

1 \(\frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
2 \(\frac{{4\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
3 \(\frac{{9\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
4 \(\frac{{36\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
KCET - 2021
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357910 The de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to

1 radius of orbit
2 perimeter of orbit
3 diameter of orbit
4 half the perimeter of orbit
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357911 An electron (of mass \(m\) ) with an initial velocity \(v = {v_0}\widehat i\) is moving in an electric field \(E = {E_0}\widehat j\). If \({\lambda _0} = h/m{v_0}\), it's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0} \sqrt{1+\dfrac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\)
2 \(\lambda_{0}\)
3 \(\frac{{{\lambda _0}}}{{\left( {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right)}}\)
4 \(\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357912 The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4 K, 2 K\) and \(K\) respectively. The de-Broglie wavelength associated with electron \(\left(\lambda_{e}\right), \alpha\)-particle \(\left(\lambda_{\alpha}\right)\) and the proton \(\left(\lambda_{p}\right)\) are as follows :

1 \(\lambda_{\alpha}>\lambda_{p}>\lambda_{e}\)
2 \(\lambda_{\alpha}=\lambda_{p} < \lambda_{e}\)
3 \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\)
4 \(\lambda_{\alpha}=\lambda_{p}>\lambda_{e}\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357913 The ratio of momentum of an electron and \(\alpha\)-particle which are accelerated from rest by a potential difference of \(100\;V\) is

1 1
2 \(\sqrt{\left(\dfrac{2 m_{e}}{m_{\alpha}}\right)}\)
3 \(\sqrt{\left(\dfrac{m_{e}}{m_{\alpha}}\right)}\)
4 \(\sqrt{\left(\dfrac{m_{e}}{2 m_{\alpha}}\right)}\)
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PHXII11:DUAL NATURE OF RADIATION AND MATTER

357909 The figure shows standing de Broglie waves due to the revolvtion of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their ususal meanings)
supporting img

1 \(\frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
2 \(\frac{{4\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
3 \(\frac{{9\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
4 \(\frac{{36\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
KCET - 2021
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357910 The de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to

1 radius of orbit
2 perimeter of orbit
3 diameter of orbit
4 half the perimeter of orbit
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357911 An electron (of mass \(m\) ) with an initial velocity \(v = {v_0}\widehat i\) is moving in an electric field \(E = {E_0}\widehat j\). If \({\lambda _0} = h/m{v_0}\), it's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0} \sqrt{1+\dfrac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\)
2 \(\lambda_{0}\)
3 \(\frac{{{\lambda _0}}}{{\left( {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right)}}\)
4 \(\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357912 The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4 K, 2 K\) and \(K\) respectively. The de-Broglie wavelength associated with electron \(\left(\lambda_{e}\right), \alpha\)-particle \(\left(\lambda_{\alpha}\right)\) and the proton \(\left(\lambda_{p}\right)\) are as follows :

1 \(\lambda_{\alpha}>\lambda_{p}>\lambda_{e}\)
2 \(\lambda_{\alpha}=\lambda_{p} < \lambda_{e}\)
3 \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\)
4 \(\lambda_{\alpha}=\lambda_{p}>\lambda_{e}\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357913 The ratio of momentum of an electron and \(\alpha\)-particle which are accelerated from rest by a potential difference of \(100\;V\) is

1 1
2 \(\sqrt{\left(\dfrac{2 m_{e}}{m_{\alpha}}\right)}\)
3 \(\sqrt{\left(\dfrac{m_{e}}{m_{\alpha}}\right)}\)
4 \(\sqrt{\left(\dfrac{m_{e}}{2 m_{\alpha}}\right)}\)
NEET DIGITAL LIBRARY
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357909 The figure shows standing de Broglie waves due to the revolvtion of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their ususal meanings)
supporting img

1 \(\frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
2 \(\frac{{4\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
3 \(\frac{{9\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
4 \(\frac{{36\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
KCET - 2021
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357910 The de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to

1 radius of orbit
2 perimeter of orbit
3 diameter of orbit
4 half the perimeter of orbit
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357911 An electron (of mass \(m\) ) with an initial velocity \(v = {v_0}\widehat i\) is moving in an electric field \(E = {E_0}\widehat j\). If \({\lambda _0} = h/m{v_0}\), it's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0} \sqrt{1+\dfrac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\)
2 \(\lambda_{0}\)
3 \(\frac{{{\lambda _0}}}{{\left( {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right)}}\)
4 \(\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357912 The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4 K, 2 K\) and \(K\) respectively. The de-Broglie wavelength associated with electron \(\left(\lambda_{e}\right), \alpha\)-particle \(\left(\lambda_{\alpha}\right)\) and the proton \(\left(\lambda_{p}\right)\) are as follows :

1 \(\lambda_{\alpha}>\lambda_{p}>\lambda_{e}\)
2 \(\lambda_{\alpha}=\lambda_{p} < \lambda_{e}\)
3 \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\)
4 \(\lambda_{\alpha}=\lambda_{p}>\lambda_{e}\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357913 The ratio of momentum of an electron and \(\alpha\)-particle which are accelerated from rest by a potential difference of \(100\;V\) is

1 1
2 \(\sqrt{\left(\dfrac{2 m_{e}}{m_{\alpha}}\right)}\)
3 \(\sqrt{\left(\dfrac{m_{e}}{m_{\alpha}}\right)}\)
4 \(\sqrt{\left(\dfrac{m_{e}}{2 m_{\alpha}}\right)}\)
Join With Us for More Updates
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357909 The figure shows standing de Broglie waves due to the revolvtion of electron in a certain orbit of hydrogen atom. Then the expression for the orbit radius is (all notations have their ususal meanings)
supporting img

1 \(\frac{{{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
2 \(\frac{{4\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
3 \(\frac{{9\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
4 \(\frac{{36\;{h^2}{\varepsilon _0}}}{{\pi m{e^2}}}\)
KCET - 2021
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357910 The de-Broglie wavelength of an electron in the first orbit of Bohr's hydrogen is equal to

1 radius of orbit
2 perimeter of orbit
3 diameter of orbit
4 half the perimeter of orbit
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357911 An electron (of mass \(m\) ) with an initial velocity \(v = {v_0}\widehat i\) is moving in an electric field \(E = {E_0}\widehat j\). If \({\lambda _0} = h/m{v_0}\), it's de Broglie wavelength at time \(t\) is given by

1 \(\lambda_{0} \sqrt{1+\dfrac{e^{2} E^{2} t^{2}}{m^{2} v_{0}^{2}}}\)
2 \(\lambda_{0}\)
3 \(\frac{{{\lambda _0}}}{{\left( {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} \right)}}\)
4 \(\frac{{{\lambda _0}}}{{\sqrt {1 + \frac{{{e^2}E_0^2{t^2}}}{{{m^2}v_0^2}}} }}\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357912 The kinetic energy of an electron, \(\alpha\)-particle and a proton are given as \(4 K, 2 K\) and \(K\) respectively. The de-Broglie wavelength associated with electron \(\left(\lambda_{e}\right), \alpha\)-particle \(\left(\lambda_{\alpha}\right)\) and the proton \(\left(\lambda_{p}\right)\) are as follows :

1 \(\lambda_{\alpha}>\lambda_{p}>\lambda_{e}\)
2 \(\lambda_{\alpha}=\lambda_{p} < \lambda_{e}\)
3 \(\lambda_{\alpha} < \lambda_{p} < \lambda_{e}\)
4 \(\lambda_{\alpha}=\lambda_{p}>\lambda_{e}\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER

357913 The ratio of momentum of an electron and \(\alpha\)-particle which are accelerated from rest by a potential difference of \(100\;V\) is

1 1
2 \(\sqrt{\left(\dfrac{2 m_{e}}{m_{\alpha}}\right)}\)
3 \(\sqrt{\left(\dfrac{m_{e}}{m_{\alpha}}\right)}\)
4 \(\sqrt{\left(\dfrac{m_{e}}{2 m_{\alpha}}\right)}\)