\(\lambda = \frac{h}{{mv}}\) \(\therefore \) De Broglie wavelength depends on mass of the particle.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357863
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 All have same wavelength
Explanation:
We know that, \(\lambda = h/p\) Given, \({\mkern 1mu} {\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,\,{p_\alpha } = {p_\beta } = {p_\gamma }\) So, all have same wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357864
If the momentum of an electron is increased by \(p_{m}\) then the de Broglie wavelength associated with it changes by \(0.5 \%\). Then find the initial momentum of the electron is
1 \(156\,pm\)
2 \(199\,pm\)
3 \(179\,pm\)
4 \(185\,pm\)
Explanation:
Let \(p\) be initial momentum of electron, \(p^{\prime}=p+p_{m}\) \(\therefore \quad \lambda^{\prime}=\lambda-\dfrac{0.5}{100} \lambda=\left(1-\dfrac{0.5}{100}\right) \lambda\) \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\dfrac{{p}}{{p}^{\prime}}\) \(\therefore\left(1-\dfrac{0.5}{100}\right)=\dfrac{p}{p+p_{m}}\) \(\therefore\left(\dfrac{99.5}{100}\right)=\dfrac{{p}}{{p}+{p}_{{m}}}\) \(\therefore \quad 99.5 \,{p}+99.5 \,{p}_{{m}}=100 \,{p}\) \(\therefore 0.5 \,{p}=99.5 \,{p}_{{m}}\) \(\therefore {p}=199 \,{p}_{{m}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357865
If the electron and proton are accelerated by same potential difference \(100\;V\), then find the ratio of their de-Broglie wavelengths \({m_e} = 9.1 \times {10^{ - 31}}\;kg\) \({m_p} = 1.6825 \times {10^{ - 27}}\;kg\)
1 43
2 \(\dfrac{1}{43}\)
3 \(\dfrac{40}{1}\)
4 \(\dfrac{1}{40}\)
Explanation:
An electron and a proton, accelerated through the same potential difference \(V\), acquire same kinetic energy \(K( = eV)\). For a particle of mass \(m\) and kinetic energy \(K\), de-Broglie wavelength: \(\lambda=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 K m}}=\dfrac{h}{\sqrt{2 e V m}}\) \(\Rightarrow \dfrac{\lambda_{e}}{\lambda_{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}} \approx 43 .\)
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357862
de Broglie wavelength depends on
1 Mass of the particle
2 Size of the particle
3 Material of the particle
4 Shape of the particle
Explanation:
\(\lambda = \frac{h}{{mv}}\) \(\therefore \) De Broglie wavelength depends on mass of the particle.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357863
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 All have same wavelength
Explanation:
We know that, \(\lambda = h/p\) Given, \({\mkern 1mu} {\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,\,{p_\alpha } = {p_\beta } = {p_\gamma }\) So, all have same wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357864
If the momentum of an electron is increased by \(p_{m}\) then the de Broglie wavelength associated with it changes by \(0.5 \%\). Then find the initial momentum of the electron is
1 \(156\,pm\)
2 \(199\,pm\)
3 \(179\,pm\)
4 \(185\,pm\)
Explanation:
Let \(p\) be initial momentum of electron, \(p^{\prime}=p+p_{m}\) \(\therefore \quad \lambda^{\prime}=\lambda-\dfrac{0.5}{100} \lambda=\left(1-\dfrac{0.5}{100}\right) \lambda\) \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\dfrac{{p}}{{p}^{\prime}}\) \(\therefore\left(1-\dfrac{0.5}{100}\right)=\dfrac{p}{p+p_{m}}\) \(\therefore\left(\dfrac{99.5}{100}\right)=\dfrac{{p}}{{p}+{p}_{{m}}}\) \(\therefore \quad 99.5 \,{p}+99.5 \,{p}_{{m}}=100 \,{p}\) \(\therefore 0.5 \,{p}=99.5 \,{p}_{{m}}\) \(\therefore {p}=199 \,{p}_{{m}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357865
If the electron and proton are accelerated by same potential difference \(100\;V\), then find the ratio of their de-Broglie wavelengths \({m_e} = 9.1 \times {10^{ - 31}}\;kg\) \({m_p} = 1.6825 \times {10^{ - 27}}\;kg\)
1 43
2 \(\dfrac{1}{43}\)
3 \(\dfrac{40}{1}\)
4 \(\dfrac{1}{40}\)
Explanation:
An electron and a proton, accelerated through the same potential difference \(V\), acquire same kinetic energy \(K( = eV)\). For a particle of mass \(m\) and kinetic energy \(K\), de-Broglie wavelength: \(\lambda=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 K m}}=\dfrac{h}{\sqrt{2 e V m}}\) \(\Rightarrow \dfrac{\lambda_{e}}{\lambda_{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}} \approx 43 .\)
\(\lambda = \frac{h}{{mv}}\) \(\therefore \) De Broglie wavelength depends on mass of the particle.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357863
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 All have same wavelength
Explanation:
We know that, \(\lambda = h/p\) Given, \({\mkern 1mu} {\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,\,{p_\alpha } = {p_\beta } = {p_\gamma }\) So, all have same wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357864
If the momentum of an electron is increased by \(p_{m}\) then the de Broglie wavelength associated with it changes by \(0.5 \%\). Then find the initial momentum of the electron is
1 \(156\,pm\)
2 \(199\,pm\)
3 \(179\,pm\)
4 \(185\,pm\)
Explanation:
Let \(p\) be initial momentum of electron, \(p^{\prime}=p+p_{m}\) \(\therefore \quad \lambda^{\prime}=\lambda-\dfrac{0.5}{100} \lambda=\left(1-\dfrac{0.5}{100}\right) \lambda\) \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\dfrac{{p}}{{p}^{\prime}}\) \(\therefore\left(1-\dfrac{0.5}{100}\right)=\dfrac{p}{p+p_{m}}\) \(\therefore\left(\dfrac{99.5}{100}\right)=\dfrac{{p}}{{p}+{p}_{{m}}}\) \(\therefore \quad 99.5 \,{p}+99.5 \,{p}_{{m}}=100 \,{p}\) \(\therefore 0.5 \,{p}=99.5 \,{p}_{{m}}\) \(\therefore {p}=199 \,{p}_{{m}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357865
If the electron and proton are accelerated by same potential difference \(100\;V\), then find the ratio of their de-Broglie wavelengths \({m_e} = 9.1 \times {10^{ - 31}}\;kg\) \({m_p} = 1.6825 \times {10^{ - 27}}\;kg\)
1 43
2 \(\dfrac{1}{43}\)
3 \(\dfrac{40}{1}\)
4 \(\dfrac{1}{40}\)
Explanation:
An electron and a proton, accelerated through the same potential difference \(V\), acquire same kinetic energy \(K( = eV)\). For a particle of mass \(m\) and kinetic energy \(K\), de-Broglie wavelength: \(\lambda=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 K m}}=\dfrac{h}{\sqrt{2 e V m}}\) \(\Rightarrow \dfrac{\lambda_{e}}{\lambda_{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}} \approx 43 .\)
\(\lambda = \frac{h}{{mv}}\) \(\therefore \) De Broglie wavelength depends on mass of the particle.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357863
If alpha, beta and gamma rays carry same momentum, which has the longest wavelength?
1 Alpha rays
2 Beta rays
3 Gamma rays
4 All have same wavelength
Explanation:
We know that, \(\lambda = h/p\) Given, \({\mkern 1mu} {\kern 1pt} \,\,\,\,\,\,\,\,\,\,\,\,{p_\alpha } = {p_\beta } = {p_\gamma }\) So, all have same wavelength.
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357864
If the momentum of an electron is increased by \(p_{m}\) then the de Broglie wavelength associated with it changes by \(0.5 \%\). Then find the initial momentum of the electron is
1 \(156\,pm\)
2 \(199\,pm\)
3 \(179\,pm\)
4 \(185\,pm\)
Explanation:
Let \(p\) be initial momentum of electron, \(p^{\prime}=p+p_{m}\) \(\therefore \quad \lambda^{\prime}=\lambda-\dfrac{0.5}{100} \lambda=\left(1-\dfrac{0.5}{100}\right) \lambda\) \(\therefore \dfrac{\lambda^{\prime}}{\lambda}=\dfrac{{p}}{{p}^{\prime}}\) \(\therefore\left(1-\dfrac{0.5}{100}\right)=\dfrac{p}{p+p_{m}}\) \(\therefore\left(\dfrac{99.5}{100}\right)=\dfrac{{p}}{{p}+{p}_{{m}}}\) \(\therefore \quad 99.5 \,{p}+99.5 \,{p}_{{m}}=100 \,{p}\) \(\therefore 0.5 \,{p}=99.5 \,{p}_{{m}}\) \(\therefore {p}=199 \,{p}_{{m}}\)
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357865
If the electron and proton are accelerated by same potential difference \(100\;V\), then find the ratio of their de-Broglie wavelengths \({m_e} = 9.1 \times {10^{ - 31}}\;kg\) \({m_p} = 1.6825 \times {10^{ - 27}}\;kg\)
1 43
2 \(\dfrac{1}{43}\)
3 \(\dfrac{40}{1}\)
4 \(\dfrac{1}{40}\)
Explanation:
An electron and a proton, accelerated through the same potential difference \(V\), acquire same kinetic energy \(K( = eV)\). For a particle of mass \(m\) and kinetic energy \(K\), de-Broglie wavelength: \(\lambda=\dfrac{h}{m v}=\dfrac{h}{\sqrt{2 K m}}=\dfrac{h}{\sqrt{2 e V m}}\) \(\Rightarrow \dfrac{\lambda_{e}}{\lambda_{p}}=\sqrt{\dfrac{m_{p}}{m_{e}}} \approx 43 .\)
