357918
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with \(1M\,eV\) energy is nearly
1 \(1.2\,nm\)
2 \(1.2 \times {10^{ - 3}}\;nm\)
3 \(1.2 \times {10^{ - 6}}\;nm\)
4 \(1.2 \times 10\;nm\)
Explanation:
Given in the equation, Energy of a photon, \(E = 1MeV \Rightarrow {10^6}eV\) Now, \(hc\) \( = 1240\,eVnm\) Now, \(E=\dfrac{h c}{\lambda}\) \( \Rightarrow \lambda = \frac{{hc}}{E} = \frac{{1240eVnm}}{{{{10}^6}eV}} = 1.24 \times {10^{ - 3}}\;nm\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357919
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 effect is
1 \(\dfrac{-e E t}{E}\)
2 \(\dfrac{-h}{e E t^{2}}\)
3 \(\dfrac{-h}{e E}\)
4 \(\dfrac{-m h}{e E t^{2}}\)
Explanation:
Here, \(u=0 ; a=\dfrac{e E}{m}\); \(\therefore \quad v = u + at = 0 + \frac{{eE}}{m}t\) (from equation of motion) de Broglie wavelength, \(\lambda = \frac{h}{{mv}} = \frac{h}{{m(eEt/m)}} = \frac{h}{{eEt}}\) Rate of change of de Broglie wavelength, \(\dfrac{d \lambda}{d t}=\dfrac{h}{e E}\left(-\dfrac{1}{t^{2}}\right)=\dfrac{-h}{e E t^{2}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357920
Calculate the momentum and de-Broglie wavelength of electrons accelerated through a potential difference of \(56\;V.\)
357921
The de - Brogile wavelength \(\lambda\) of a particle
1 Is proportional to mass
2 Is proportional to impulse
3 Is inversely proportional to impulse
4 Does not depend on impule
Explanation:
de - Brogile wavelength \((\lambda)\) of a particle is related to its momentum (\(p\)) as \(\lambda=\dfrac{h}{p}\) where, \(h\) is Planck's constant \({\text{ or }}\lambda = \frac{h}{{mv}}\quad [\because p = mv\} \) \(\quad \lambda = \frac{h}{{m\frac{v}{t}.t}} = \frac{h}{{ma \cdot t}} = \frac{h}{{F \cdot t}}\) \([\because F = ma]\) \((\lambda )\) is inversely proportional to impulse \((F)\)
357918
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with \(1M\,eV\) energy is nearly
1 \(1.2\,nm\)
2 \(1.2 \times {10^{ - 3}}\;nm\)
3 \(1.2 \times {10^{ - 6}}\;nm\)
4 \(1.2 \times 10\;nm\)
Explanation:
Given in the equation, Energy of a photon, \(E = 1MeV \Rightarrow {10^6}eV\) Now, \(hc\) \( = 1240\,eVnm\) Now, \(E=\dfrac{h c}{\lambda}\) \( \Rightarrow \lambda = \frac{{hc}}{E} = \frac{{1240eVnm}}{{{{10}^6}eV}} = 1.24 \times {10^{ - 3}}\;nm\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357919
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 effect is
1 \(\dfrac{-e E t}{E}\)
2 \(\dfrac{-h}{e E t^{2}}\)
3 \(\dfrac{-h}{e E}\)
4 \(\dfrac{-m h}{e E t^{2}}\)
Explanation:
Here, \(u=0 ; a=\dfrac{e E}{m}\); \(\therefore \quad v = u + at = 0 + \frac{{eE}}{m}t\) (from equation of motion) de Broglie wavelength, \(\lambda = \frac{h}{{mv}} = \frac{h}{{m(eEt/m)}} = \frac{h}{{eEt}}\) Rate of change of de Broglie wavelength, \(\dfrac{d \lambda}{d t}=\dfrac{h}{e E}\left(-\dfrac{1}{t^{2}}\right)=\dfrac{-h}{e E t^{2}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357920
Calculate the momentum and de-Broglie wavelength of electrons accelerated through a potential difference of \(56\;V.\)
357921
The de - Brogile wavelength \(\lambda\) of a particle
1 Is proportional to mass
2 Is proportional to impulse
3 Is inversely proportional to impulse
4 Does not depend on impule
Explanation:
de - Brogile wavelength \((\lambda)\) of a particle is related to its momentum (\(p\)) as \(\lambda=\dfrac{h}{p}\) where, \(h\) is Planck's constant \({\text{ or }}\lambda = \frac{h}{{mv}}\quad [\because p = mv\} \) \(\quad \lambda = \frac{h}{{m\frac{v}{t}.t}} = \frac{h}{{ma \cdot t}} = \frac{h}{{F \cdot t}}\) \([\because F = ma]\) \((\lambda )\) is inversely proportional to impulse \((F)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357918
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with \(1M\,eV\) energy is nearly
1 \(1.2\,nm\)
2 \(1.2 \times {10^{ - 3}}\;nm\)
3 \(1.2 \times {10^{ - 6}}\;nm\)
4 \(1.2 \times 10\;nm\)
Explanation:
Given in the equation, Energy of a photon, \(E = 1MeV \Rightarrow {10^6}eV\) Now, \(hc\) \( = 1240\,eVnm\) Now, \(E=\dfrac{h c}{\lambda}\) \( \Rightarrow \lambda = \frac{{hc}}{E} = \frac{{1240eVnm}}{{{{10}^6}eV}} = 1.24 \times {10^{ - 3}}\;nm\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357919
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 effect is
1 \(\dfrac{-e E t}{E}\)
2 \(\dfrac{-h}{e E t^{2}}\)
3 \(\dfrac{-h}{e E}\)
4 \(\dfrac{-m h}{e E t^{2}}\)
Explanation:
Here, \(u=0 ; a=\dfrac{e E}{m}\); \(\therefore \quad v = u + at = 0 + \frac{{eE}}{m}t\) (from equation of motion) de Broglie wavelength, \(\lambda = \frac{h}{{mv}} = \frac{h}{{m(eEt/m)}} = \frac{h}{{eEt}}\) Rate of change of de Broglie wavelength, \(\dfrac{d \lambda}{d t}=\dfrac{h}{e E}\left(-\dfrac{1}{t^{2}}\right)=\dfrac{-h}{e E t^{2}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357920
Calculate the momentum and de-Broglie wavelength of electrons accelerated through a potential difference of \(56\;V.\)
357921
The de - Brogile wavelength \(\lambda\) of a particle
1 Is proportional to mass
2 Is proportional to impulse
3 Is inversely proportional to impulse
4 Does not depend on impule
Explanation:
de - Brogile wavelength \((\lambda)\) of a particle is related to its momentum (\(p\)) as \(\lambda=\dfrac{h}{p}\) where, \(h\) is Planck's constant \({\text{ or }}\lambda = \frac{h}{{mv}}\quad [\because p = mv\} \) \(\quad \lambda = \frac{h}{{m\frac{v}{t}.t}} = \frac{h}{{ma \cdot t}} = \frac{h}{{F \cdot t}}\) \([\because F = ma]\) \((\lambda )\) is inversely proportional to impulse \((F)\)
357918
The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with \(1M\,eV\) energy is nearly
1 \(1.2\,nm\)
2 \(1.2 \times {10^{ - 3}}\;nm\)
3 \(1.2 \times {10^{ - 6}}\;nm\)
4 \(1.2 \times 10\;nm\)
Explanation:
Given in the equation, Energy of a photon, \(E = 1MeV \Rightarrow {10^6}eV\) Now, \(hc\) \( = 1240\,eVnm\) Now, \(E=\dfrac{h c}{\lambda}\) \( \Rightarrow \lambda = \frac{{hc}}{E} = \frac{{1240eVnm}}{{{{10}^6}eV}} = 1.24 \times {10^{ - 3}}\;nm\)
NCERT Exempler
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357919
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 effect is
1 \(\dfrac{-e E t}{E}\)
2 \(\dfrac{-h}{e E t^{2}}\)
3 \(\dfrac{-h}{e E}\)
4 \(\dfrac{-m h}{e E t^{2}}\)
Explanation:
Here, \(u=0 ; a=\dfrac{e E}{m}\); \(\therefore \quad v = u + at = 0 + \frac{{eE}}{m}t\) (from equation of motion) de Broglie wavelength, \(\lambda = \frac{h}{{mv}} = \frac{h}{{m(eEt/m)}} = \frac{h}{{eEt}}\) Rate of change of de Broglie wavelength, \(\dfrac{d \lambda}{d t}=\dfrac{h}{e E}\left(-\dfrac{1}{t^{2}}\right)=\dfrac{-h}{e E t^{2}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357920
Calculate the momentum and de-Broglie wavelength of electrons accelerated through a potential difference of \(56\;V.\)
357921
The de - Brogile wavelength \(\lambda\) of a particle
1 Is proportional to mass
2 Is proportional to impulse
3 Is inversely proportional to impulse
4 Does not depend on impule
Explanation:
de - Brogile wavelength \((\lambda)\) of a particle is related to its momentum (\(p\)) as \(\lambda=\dfrac{h}{p}\) where, \(h\) is Planck's constant \({\text{ or }}\lambda = \frac{h}{{mv}}\quad [\because p = mv\} \) \(\quad \lambda = \frac{h}{{m\frac{v}{t}.t}} = \frac{h}{{ma \cdot t}} = \frac{h}{{F \cdot t}}\) \([\because F = ma]\) \((\lambda )\) is inversely proportional to impulse \((F)\)