142562
The following particles particles are moving with the same velocity, then maximum deBroglie wavelength will be for
1 proton
2 $\alpha$-particle
3 neutron
4 $\beta$-particle
Explanation:
D We know that $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Where, $\mathrm{m}, \mathrm{v}$ are the mass and velocity. All the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength. $\beta$ particles has the lowest mass and therefore it has maximum wavelength.
AIPMT - 2002
Dual nature of radiation and Matter
142564
The wavelength associated with an electron, accelerated through a potential difference of $100 \mathrm{~V}$, is of the order of
142565
The de-Broglie wave corresponding to a particle of mass $m$ and velocity $v$ has a wavelength associated with it
1 $\frac{\mathrm{h}}{\mathrm{mv}}$
2 hmv
3 $\frac{\mathrm{mh}}{\mathrm{v}}$
4 $\frac{\mathrm{m}}{\mathrm{hv}}$
Explanation:
A de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\mathrm{h}=$ planck's constant $\mathrm{p} =\text { momentum }$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$
AIPMT - 1989
Dual nature of radiation and Matter
142575
The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is: c-velocity of light, h-Planck's constant
1 \(\mathrm{h}\)
2 \(\mathrm{c}\)
3 \(\frac{1}{\mathrm{~h}}\)
4 \(\frac{1}{\mathrm{c}}\)
Explanation:
B Given that, \(\lambda_{\text {ph }}=\lambda_{\mathrm{e}}\) Now, \(\quad \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{c}}}\) \(\mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\mathrm{e}}}\) And, \(\quad \mathrm{E}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}\) From equation (i) and (ii) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}} \cdot \frac{\lambda_{\mathrm{e}}}{\mathrm{h}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{c} \cdot \lambda_{\mathrm{e}}}{\lambda_{\mathrm{ph}}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\mathrm{c}\)\(\left(\right.\) given,\(\left.\lambda_{\mathrm{e}}=\lambda_{\mathrm{ph}}\right)\)
142562
The following particles particles are moving with the same velocity, then maximum deBroglie wavelength will be for
1 proton
2 $\alpha$-particle
3 neutron
4 $\beta$-particle
Explanation:
D We know that $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Where, $\mathrm{m}, \mathrm{v}$ are the mass and velocity. All the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength. $\beta$ particles has the lowest mass and therefore it has maximum wavelength.
AIPMT - 2002
Dual nature of radiation and Matter
142564
The wavelength associated with an electron, accelerated through a potential difference of $100 \mathrm{~V}$, is of the order of
142565
The de-Broglie wave corresponding to a particle of mass $m$ and velocity $v$ has a wavelength associated with it
1 $\frac{\mathrm{h}}{\mathrm{mv}}$
2 hmv
3 $\frac{\mathrm{mh}}{\mathrm{v}}$
4 $\frac{\mathrm{m}}{\mathrm{hv}}$
Explanation:
A de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\mathrm{h}=$ planck's constant $\mathrm{p} =\text { momentum }$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$
AIPMT - 1989
Dual nature of radiation and Matter
142575
The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is: c-velocity of light, h-Planck's constant
1 \(\mathrm{h}\)
2 \(\mathrm{c}\)
3 \(\frac{1}{\mathrm{~h}}\)
4 \(\frac{1}{\mathrm{c}}\)
Explanation:
B Given that, \(\lambda_{\text {ph }}=\lambda_{\mathrm{e}}\) Now, \(\quad \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{c}}}\) \(\mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\mathrm{e}}}\) And, \(\quad \mathrm{E}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}\) From equation (i) and (ii) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}} \cdot \frac{\lambda_{\mathrm{e}}}{\mathrm{h}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{c} \cdot \lambda_{\mathrm{e}}}{\lambda_{\mathrm{ph}}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\mathrm{c}\)\(\left(\right.\) given,\(\left.\lambda_{\mathrm{e}}=\lambda_{\mathrm{ph}}\right)\)
142562
The following particles particles are moving with the same velocity, then maximum deBroglie wavelength will be for
1 proton
2 $\alpha$-particle
3 neutron
4 $\beta$-particle
Explanation:
D We know that $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Where, $\mathrm{m}, \mathrm{v}$ are the mass and velocity. All the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength. $\beta$ particles has the lowest mass and therefore it has maximum wavelength.
AIPMT - 2002
Dual nature of radiation and Matter
142564
The wavelength associated with an electron, accelerated through a potential difference of $100 \mathrm{~V}$, is of the order of
142565
The de-Broglie wave corresponding to a particle of mass $m$ and velocity $v$ has a wavelength associated with it
1 $\frac{\mathrm{h}}{\mathrm{mv}}$
2 hmv
3 $\frac{\mathrm{mh}}{\mathrm{v}}$
4 $\frac{\mathrm{m}}{\mathrm{hv}}$
Explanation:
A de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\mathrm{h}=$ planck's constant $\mathrm{p} =\text { momentum }$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$
AIPMT - 1989
Dual nature of radiation and Matter
142575
The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is: c-velocity of light, h-Planck's constant
1 \(\mathrm{h}\)
2 \(\mathrm{c}\)
3 \(\frac{1}{\mathrm{~h}}\)
4 \(\frac{1}{\mathrm{c}}\)
Explanation:
B Given that, \(\lambda_{\text {ph }}=\lambda_{\mathrm{e}}\) Now, \(\quad \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{c}}}\) \(\mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\mathrm{e}}}\) And, \(\quad \mathrm{E}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}\) From equation (i) and (ii) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}} \cdot \frac{\lambda_{\mathrm{e}}}{\mathrm{h}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{c} \cdot \lambda_{\mathrm{e}}}{\lambda_{\mathrm{ph}}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\mathrm{c}\)\(\left(\right.\) given,\(\left.\lambda_{\mathrm{e}}=\lambda_{\mathrm{ph}}\right)\)
142562
The following particles particles are moving with the same velocity, then maximum deBroglie wavelength will be for
1 proton
2 $\alpha$-particle
3 neutron
4 $\beta$-particle
Explanation:
D We know that $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ Where, $\mathrm{m}, \mathrm{v}$ are the mass and velocity. All the particles are moving with same velocity, the particle with least mass will have maximum de-Broglie wavelength. $\beta$ particles has the lowest mass and therefore it has maximum wavelength.
AIPMT - 2002
Dual nature of radiation and Matter
142564
The wavelength associated with an electron, accelerated through a potential difference of $100 \mathrm{~V}$, is of the order of
142565
The de-Broglie wave corresponding to a particle of mass $m$ and velocity $v$ has a wavelength associated with it
1 $\frac{\mathrm{h}}{\mathrm{mv}}$
2 hmv
3 $\frac{\mathrm{mh}}{\mathrm{v}}$
4 $\frac{\mathrm{m}}{\mathrm{hv}}$
Explanation:
A de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Where $\mathrm{h}=$ planck's constant $\mathrm{p} =\text { momentum }$ $\lambda =\frac{\mathrm{h}}{\mathrm{mv}} \quad(\because \mathrm{p}=\mathrm{mv})$
AIPMT - 1989
Dual nature of radiation and Matter
142575
The de-Broglie wavelength of an electron and the wavelength of a photon are the same. The ratio between the energy of that photon and the momentum of that electron is: c-velocity of light, h-Planck's constant
1 \(\mathrm{h}\)
2 \(\mathrm{c}\)
3 \(\frac{1}{\mathrm{~h}}\)
4 \(\frac{1}{\mathrm{c}}\)
Explanation:
B Given that, \(\lambda_{\text {ph }}=\lambda_{\mathrm{e}}\) Now, \(\quad \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{c}}}\) \(\mathrm{p}_{\mathrm{e}}=\frac{\mathrm{h}}{\lambda_{\mathrm{e}}}\) And, \(\quad \mathrm{E}_{\mathrm{ph}}=\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}}\) From equation (i) and (ii) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{hc}}{\lambda_{\mathrm{ph}}} \cdot \frac{\lambda_{\mathrm{e}}}{\mathrm{h}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\frac{\mathrm{c} \cdot \lambda_{\mathrm{e}}}{\lambda_{\mathrm{ph}}}\) \(\frac{\mathrm{E}_{\mathrm{ph}}}{\mathrm{p}_{\mathrm{e}}} =\mathrm{c}\)\(\left(\right.\) given,\(\left.\lambda_{\mathrm{e}}=\lambda_{\mathrm{ph}}\right)\)
