142493
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, $h=$ Plank's constant $)$
1 $\mathrm{h}$
2 c
3 $\frac{1}{\mathrm{~h}}$
4 $\frac{1}{\mathrm{c}}$
Explanation:
B de-Broglie wavelength of electron equal to wavelength of proton, $\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}=\frac{\mathrm{hc}}{\mathrm{p}_{\mathrm{e}} \mathrm{c}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}} \quad\left(\because \mathrm{E}_{\mathrm{p}}=\mathrm{p}_{\mathrm{e}} \mathrm{c}\right)$ $\text { So, } \quad \frac{E_{p}}{p_{e}}=\frac{h c}{h}=c$
JCECE-2014
Dual nature of radiation and Matter
142496
For a particle of mass $m$ enclosed in a onedimensional box of length $L$, the de-Broglie concept would lead to stationary waves, with nodes at the two ends. The energy values allowed for such a system (with $n$ as integer) will be
4 $\frac{\mathrm{h}^{2}}{4 m L^{2}} \mathrm{n}^{2}$
Explanation:
A In an one dimensional, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{2 \mathrm{~L}}{\mathrm{n}} \text { and } \mathrm{p}=\frac{\mathrm{nh}}{2 \mathrm{~L}}$ Energy for a system $(E)=\frac{p^{2}}{2 m}$ $\mathrm{E}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \mathrm{~L}^{2} \mathrm{~m}}$
JCECE-2011
Dual nature of radiation and Matter
142497
Photon and electron are given energy $\left(10^{-2} \mathrm{~J}\right)$. Wavelengths associated with photon and electron are $\lambda_{\text {ph }}$ and $\lambda_{\text {el }}$ then, correct statement will be
A Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron. de-Broglie wavelength associated with energy- $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Then, $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ Therefore, $\lambda_{\mathrm{ph}}>\lambda_{\mathrm{el}}$
JCECE-2010
Dual nature of radiation and Matter
142499
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D If $p$ is the momentum of the particle the wavelength of the wave associated with it, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Since, alpha, beta and gamma rays carry the same momentum, so they will have the same wavelength.
142493
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, $h=$ Plank's constant $)$
1 $\mathrm{h}$
2 c
3 $\frac{1}{\mathrm{~h}}$
4 $\frac{1}{\mathrm{c}}$
Explanation:
B de-Broglie wavelength of electron equal to wavelength of proton, $\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}=\frac{\mathrm{hc}}{\mathrm{p}_{\mathrm{e}} \mathrm{c}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}} \quad\left(\because \mathrm{E}_{\mathrm{p}}=\mathrm{p}_{\mathrm{e}} \mathrm{c}\right)$ $\text { So, } \quad \frac{E_{p}}{p_{e}}=\frac{h c}{h}=c$
JCECE-2014
Dual nature of radiation and Matter
142496
For a particle of mass $m$ enclosed in a onedimensional box of length $L$, the de-Broglie concept would lead to stationary waves, with nodes at the two ends. The energy values allowed for such a system (with $n$ as integer) will be
4 $\frac{\mathrm{h}^{2}}{4 m L^{2}} \mathrm{n}^{2}$
Explanation:
A In an one dimensional, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{2 \mathrm{~L}}{\mathrm{n}} \text { and } \mathrm{p}=\frac{\mathrm{nh}}{2 \mathrm{~L}}$ Energy for a system $(E)=\frac{p^{2}}{2 m}$ $\mathrm{E}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \mathrm{~L}^{2} \mathrm{~m}}$
JCECE-2011
Dual nature of radiation and Matter
142497
Photon and electron are given energy $\left(10^{-2} \mathrm{~J}\right)$. Wavelengths associated with photon and electron are $\lambda_{\text {ph }}$ and $\lambda_{\text {el }}$ then, correct statement will be
A Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron. de-Broglie wavelength associated with energy- $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Then, $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ Therefore, $\lambda_{\mathrm{ph}}>\lambda_{\mathrm{el}}$
JCECE-2010
Dual nature of radiation and Matter
142499
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D If $p$ is the momentum of the particle the wavelength of the wave associated with it, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Since, alpha, beta and gamma rays carry the same momentum, so they will have the same wavelength.
142493
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, $h=$ Plank's constant $)$
1 $\mathrm{h}$
2 c
3 $\frac{1}{\mathrm{~h}}$
4 $\frac{1}{\mathrm{c}}$
Explanation:
B de-Broglie wavelength of electron equal to wavelength of proton, $\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}=\frac{\mathrm{hc}}{\mathrm{p}_{\mathrm{e}} \mathrm{c}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}} \quad\left(\because \mathrm{E}_{\mathrm{p}}=\mathrm{p}_{\mathrm{e}} \mathrm{c}\right)$ $\text { So, } \quad \frac{E_{p}}{p_{e}}=\frac{h c}{h}=c$
JCECE-2014
Dual nature of radiation and Matter
142496
For a particle of mass $m$ enclosed in a onedimensional box of length $L$, the de-Broglie concept would lead to stationary waves, with nodes at the two ends. The energy values allowed for such a system (with $n$ as integer) will be
4 $\frac{\mathrm{h}^{2}}{4 m L^{2}} \mathrm{n}^{2}$
Explanation:
A In an one dimensional, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{2 \mathrm{~L}}{\mathrm{n}} \text { and } \mathrm{p}=\frac{\mathrm{nh}}{2 \mathrm{~L}}$ Energy for a system $(E)=\frac{p^{2}}{2 m}$ $\mathrm{E}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \mathrm{~L}^{2} \mathrm{~m}}$
JCECE-2011
Dual nature of radiation and Matter
142497
Photon and electron are given energy $\left(10^{-2} \mathrm{~J}\right)$. Wavelengths associated with photon and electron are $\lambda_{\text {ph }}$ and $\lambda_{\text {el }}$ then, correct statement will be
A Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron. de-Broglie wavelength associated with energy- $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Then, $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ Therefore, $\lambda_{\mathrm{ph}}>\lambda_{\mathrm{el}}$
JCECE-2010
Dual nature of radiation and Matter
142499
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D If $p$ is the momentum of the particle the wavelength of the wave associated with it, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Since, alpha, beta and gamma rays carry the same momentum, so they will have the same wavelength.
142493
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, $h=$ Plank's constant $)$
1 $\mathrm{h}$
2 c
3 $\frac{1}{\mathrm{~h}}$
4 $\frac{1}{\mathrm{c}}$
Explanation:
B de-Broglie wavelength of electron equal to wavelength of proton, $\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}}=\frac{\mathrm{hc}}{\mathrm{p}_{\mathrm{e}} \mathrm{c}}=\frac{\mathrm{hc}}{\mathrm{E}_{\mathrm{p}}} \quad\left(\because \mathrm{E}_{\mathrm{p}}=\mathrm{p}_{\mathrm{e}} \mathrm{c}\right)$ $\text { So, } \quad \frac{E_{p}}{p_{e}}=\frac{h c}{h}=c$
JCECE-2014
Dual nature of radiation and Matter
142496
For a particle of mass $m$ enclosed in a onedimensional box of length $L$, the de-Broglie concept would lead to stationary waves, with nodes at the two ends. The energy values allowed for such a system (with $n$ as integer) will be
4 $\frac{\mathrm{h}^{2}}{4 m L^{2}} \mathrm{n}^{2}$
Explanation:
A In an one dimensional, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{2 \mathrm{~L}}{\mathrm{n}} \text { and } \mathrm{p}=\frac{\mathrm{nh}}{2 \mathrm{~L}}$ Energy for a system $(E)=\frac{p^{2}}{2 m}$ $\mathrm{E}=\frac{\mathrm{n}^{2} \mathrm{~h}^{2}}{8 \mathrm{~L}^{2} \mathrm{~m}}$
JCECE-2011
Dual nature of radiation and Matter
142497
Photon and electron are given energy $\left(10^{-2} \mathrm{~J}\right)$. Wavelengths associated with photon and electron are $\lambda_{\text {ph }}$ and $\lambda_{\text {el }}$ then, correct statement will be
A Wavelength of photon will be greater than that of electron because mass of photon is less than that of electron. de-Broglie wavelength associated with energy- $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Then, $\quad \lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ Therefore, $\lambda_{\mathrm{ph}}>\lambda_{\mathrm{el}}$
JCECE-2010
Dual nature of radiation and Matter
142499
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 None, all have same wavelength
Explanation:
D If $p$ is the momentum of the particle the wavelength of the wave associated with it, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ Since, alpha, beta and gamma rays carry the same momentum, so they will have the same wavelength.
