NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXII11:DUAL NATURE OF RADIATION AND MATTER
357854
The ratio of de-Broglie wavelength of an \(\alpha\) particle and a proton accelerated from rest by the same potential is \(\dfrac{1}{\sqrt{m}}\), the value of \(m\) is
1 16
2 2
3 4
4 8
Explanation:
Here: \(m_{\alpha}=4 m_{P}, q_{\alpha}=2 q_{P}\), \(V=\) Potential difference. \(\lambda=\dfrac{h}{\sqrt{2 m q V}}\) so, \(\dfrac{\lambda_{\alpha}}{\lambda_{P}}=\sqrt{\dfrac{2 m_{P} q_{P} V}{2 m_{\alpha} q_{\alpha} V}}=\sqrt{\dfrac{m_{P} \cdot q_{P}}{4 m_{P} \times 2 q_{P}}}=\dfrac{1}{\sqrt{8}}\) So, \(m=8\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357855
If the momentum of an electron is changed by \(\Delta p\), then the de-Broglie wavelength associated with its changes by \(0.5 \%\). The initial momentum of electron will be:
357856
Proton \((P)\) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, \(m_{P}=1849 m_{e}\) ).
1 \(1: 43\)
2 \(43: 1\)
3 \(1: 1849\)
4 \(1: 1\)
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}\), where to is the linear momentum of particle. Given \(: \dfrac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=1\) \(\dfrac{h / p_{\text {proton }}}{h / p_{\text {electron }}}=1 \Rightarrow \dfrac{p_{\text {proton }}}{p_{\text {electron }}}=1\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357857
What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 \(volt\)?
1 \(0.1227\mathop A\limits^o \)
2 \(12.27\mathop A\limits^o \)
3 \(0.001227\mathop A\limits^o \)
4 \(1.227\mathop A\limits^o \)
Explanation:
de Broglie wavelength of the electron accelerated through a potential difference of \(V\) volt is \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o \) For \(V = 100\,volt\) \(\lambda = \frac{{12.27}}{{\sqrt {100} }}\mathop A\limits^o = 1.227\mathop A\limits^o \)
357854
The ratio of de-Broglie wavelength of an \(\alpha\) particle and a proton accelerated from rest by the same potential is \(\dfrac{1}{\sqrt{m}}\), the value of \(m\) is
1 16
2 2
3 4
4 8
Explanation:
Here: \(m_{\alpha}=4 m_{P}, q_{\alpha}=2 q_{P}\), \(V=\) Potential difference. \(\lambda=\dfrac{h}{\sqrt{2 m q V}}\) so, \(\dfrac{\lambda_{\alpha}}{\lambda_{P}}=\sqrt{\dfrac{2 m_{P} q_{P} V}{2 m_{\alpha} q_{\alpha} V}}=\sqrt{\dfrac{m_{P} \cdot q_{P}}{4 m_{P} \times 2 q_{P}}}=\dfrac{1}{\sqrt{8}}\) So, \(m=8\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357855
If the momentum of an electron is changed by \(\Delta p\), then the de-Broglie wavelength associated with its changes by \(0.5 \%\). The initial momentum of electron will be:
357856
Proton \((P)\) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, \(m_{P}=1849 m_{e}\) ).
1 \(1: 43\)
2 \(43: 1\)
3 \(1: 1849\)
4 \(1: 1\)
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}\), where to is the linear momentum of particle. Given \(: \dfrac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=1\) \(\dfrac{h / p_{\text {proton }}}{h / p_{\text {electron }}}=1 \Rightarrow \dfrac{p_{\text {proton }}}{p_{\text {electron }}}=1\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357857
What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 \(volt\)?
1 \(0.1227\mathop A\limits^o \)
2 \(12.27\mathop A\limits^o \)
3 \(0.001227\mathop A\limits^o \)
4 \(1.227\mathop A\limits^o \)
Explanation:
de Broglie wavelength of the electron accelerated through a potential difference of \(V\) volt is \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o \) For \(V = 100\,volt\) \(\lambda = \frac{{12.27}}{{\sqrt {100} }}\mathop A\limits^o = 1.227\mathop A\limits^o \)
357854
The ratio of de-Broglie wavelength of an \(\alpha\) particle and a proton accelerated from rest by the same potential is \(\dfrac{1}{\sqrt{m}}\), the value of \(m\) is
1 16
2 2
3 4
4 8
Explanation:
Here: \(m_{\alpha}=4 m_{P}, q_{\alpha}=2 q_{P}\), \(V=\) Potential difference. \(\lambda=\dfrac{h}{\sqrt{2 m q V}}\) so, \(\dfrac{\lambda_{\alpha}}{\lambda_{P}}=\sqrt{\dfrac{2 m_{P} q_{P} V}{2 m_{\alpha} q_{\alpha} V}}=\sqrt{\dfrac{m_{P} \cdot q_{P}}{4 m_{P} \times 2 q_{P}}}=\dfrac{1}{\sqrt{8}}\) So, \(m=8\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357855
If the momentum of an electron is changed by \(\Delta p\), then the de-Broglie wavelength associated with its changes by \(0.5 \%\). The initial momentum of electron will be:
357856
Proton \((P)\) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, \(m_{P}=1849 m_{e}\) ).
1 \(1: 43\)
2 \(43: 1\)
3 \(1: 1849\)
4 \(1: 1\)
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}\), where to is the linear momentum of particle. Given \(: \dfrac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=1\) \(\dfrac{h / p_{\text {proton }}}{h / p_{\text {electron }}}=1 \Rightarrow \dfrac{p_{\text {proton }}}{p_{\text {electron }}}=1\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357857
What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 \(volt\)?
1 \(0.1227\mathop A\limits^o \)
2 \(12.27\mathop A\limits^o \)
3 \(0.001227\mathop A\limits^o \)
4 \(1.227\mathop A\limits^o \)
Explanation:
de Broglie wavelength of the electron accelerated through a potential difference of \(V\) volt is \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o \) For \(V = 100\,volt\) \(\lambda = \frac{{12.27}}{{\sqrt {100} }}\mathop A\limits^o = 1.227\mathop A\limits^o \)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357854
The ratio of de-Broglie wavelength of an \(\alpha\) particle and a proton accelerated from rest by the same potential is \(\dfrac{1}{\sqrt{m}}\), the value of \(m\) is
1 16
2 2
3 4
4 8
Explanation:
Here: \(m_{\alpha}=4 m_{P}, q_{\alpha}=2 q_{P}\), \(V=\) Potential difference. \(\lambda=\dfrac{h}{\sqrt{2 m q V}}\) so, \(\dfrac{\lambda_{\alpha}}{\lambda_{P}}=\sqrt{\dfrac{2 m_{P} q_{P} V}{2 m_{\alpha} q_{\alpha} V}}=\sqrt{\dfrac{m_{P} \cdot q_{P}}{4 m_{P} \times 2 q_{P}}}=\dfrac{1}{\sqrt{8}}\) So, \(m=8\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357855
If the momentum of an electron is changed by \(\Delta p\), then the de-Broglie wavelength associated with its changes by \(0.5 \%\). The initial momentum of electron will be:
357856
Proton \((P)\) and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, \(m_{P}=1849 m_{e}\) ).
1 \(1: 43\)
2 \(43: 1\)
3 \(1: 1849\)
4 \(1: 1\)
Explanation:
De-Broglie wavelength, \(\lambda=\dfrac{h}{p}\), where to is the linear momentum of particle. Given \(: \dfrac{\lambda_{\text {proton }}}{\lambda_{\text {electron }}}=1\) \(\dfrac{h / p_{\text {proton }}}{h / p_{\text {electron }}}=1 \Rightarrow \dfrac{p_{\text {proton }}}{p_{\text {electron }}}=1\)
JEE - 2023
PHXII11:DUAL NATURE OF RADIATION AND MATTER
357857
What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 \(volt\)?
1 \(0.1227\mathop A\limits^o \)
2 \(12.27\mathop A\limits^o \)
3 \(0.001227\mathop A\limits^o \)
4 \(1.227\mathop A\limits^o \)
Explanation:
de Broglie wavelength of the electron accelerated through a potential difference of \(V\) volt is \(\lambda = \frac{{12.27}}{{\sqrt V }}\mathop A\limits^o \) For \(V = 100\,volt\) \(\lambda = \frac{{12.27}}{{\sqrt {100} }}\mathop A\limits^o = 1.227\mathop A\limits^o \)