142619
An element with atomic number \(Z=11\) emits \(\mathrm{E}_\alpha\) - X-ray of wavelength \(\lambda\). The atomic number which emits \(\mathrm{E}_\alpha-\mathrm{X}\)-ray of wavelength \(4 \lambda\) is
1 4
2 6
3 11
4 44
Explanation:
B Given, Atomic \(\operatorname{Number}\left(Z_1\right)=1\) Wavelength \(\left(\lambda_1\right)=\lambda\) Wavelength of second element \(\left(\lambda_2\right)=4 \lambda\). Then atomic number of second element \(\left(Z_2\right)=\) ? From Moseley's law- \(\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})\) \(\mathrm{f}=\mathrm{a}^2(Z-b)^2\) \(\frac{\lambda_1}{\lambda_2}=\frac{\left(Z_2-1\right)}{\left(Z_1-1\right)} \quad\left(\because \mathrm{f}=\frac{1}{\lambda}\right)\) \(\frac{\lambda}{4 \lambda}=\frac{\left(Z_2-1\right)^2}{(11-1)^2}\) \(\frac{100}{4}=\left(Z_2-1\right)^2\) \(\mathrm{Z}_2-1=5\) \(Z_2=6\)
VITEEE-2013
Dual nature of radiation and Matter
142621
The graph between the square root of the frequency of a specific line of characteristic spectrum of \(\mathrm{X}\) - ray and the atomic number of the target will be
1
2
3
4
Explanation:
B Relationship between frequency of characteristic spectrum of X-ray and atomic number - \(v=\mathrm{a}^2(\mathrm{Z}-\mathrm{b})^2\) Where, \(v=\) frequency \(\mathrm{Z}=\) atomic number a \& b are constant \(\sqrt{v}=a Z-a b\) We know, straight line equation \(\mathrm{Y}=\mathrm{mx}+\mathrm{c}\) Comparing equation (i) \& (ii), we get graph like this
VITEEE-2012
Dual nature of radiation and Matter
142625
If an X-ray tube is operated at \(15 \mathrm{kV}\), then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be
142626
When a beam of accelerated electrons hits a target, a continuous \(\mathrm{X}\)-rays spectrum is emitted from the target. Which of the following wavelength is absent in the \(X\)-ray spectrum, if the X-ray tube is operating at \(40000 \mathrm{~V}\) ?
1 \(0.5 \AA\)
2 \(1.0 \AA\)
3 \(0.25 \AA\)
4 \(1.5 \AA\)
Explanation:
C X-ray tube operating (V) \(=40000 \mathrm{~V}\) We know, \(\lambda_{\text {sutoff }} =\frac{\mathrm{hc}}{\mathrm{eV}}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 40000}\) \(\lambda_{\text {sutofi }}=0.31 \AA\) \(\lambda>0.31 \AA\) wavelength is absent in the \(\mathrm{X}\)-ray spectrum.
142619
An element with atomic number \(Z=11\) emits \(\mathrm{E}_\alpha\) - X-ray of wavelength \(\lambda\). The atomic number which emits \(\mathrm{E}_\alpha-\mathrm{X}\)-ray of wavelength \(4 \lambda\) is
1 4
2 6
3 11
4 44
Explanation:
B Given, Atomic \(\operatorname{Number}\left(Z_1\right)=1\) Wavelength \(\left(\lambda_1\right)=\lambda\) Wavelength of second element \(\left(\lambda_2\right)=4 \lambda\). Then atomic number of second element \(\left(Z_2\right)=\) ? From Moseley's law- \(\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})\) \(\mathrm{f}=\mathrm{a}^2(Z-b)^2\) \(\frac{\lambda_1}{\lambda_2}=\frac{\left(Z_2-1\right)}{\left(Z_1-1\right)} \quad\left(\because \mathrm{f}=\frac{1}{\lambda}\right)\) \(\frac{\lambda}{4 \lambda}=\frac{\left(Z_2-1\right)^2}{(11-1)^2}\) \(\frac{100}{4}=\left(Z_2-1\right)^2\) \(\mathrm{Z}_2-1=5\) \(Z_2=6\)
VITEEE-2013
Dual nature of radiation and Matter
142621
The graph between the square root of the frequency of a specific line of characteristic spectrum of \(\mathrm{X}\) - ray and the atomic number of the target will be
1
2
3
4
Explanation:
B Relationship between frequency of characteristic spectrum of X-ray and atomic number - \(v=\mathrm{a}^2(\mathrm{Z}-\mathrm{b})^2\) Where, \(v=\) frequency \(\mathrm{Z}=\) atomic number a \& b are constant \(\sqrt{v}=a Z-a b\) We know, straight line equation \(\mathrm{Y}=\mathrm{mx}+\mathrm{c}\) Comparing equation (i) \& (ii), we get graph like this
VITEEE-2012
Dual nature of radiation and Matter
142625
If an X-ray tube is operated at \(15 \mathrm{kV}\), then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be
142626
When a beam of accelerated electrons hits a target, a continuous \(\mathrm{X}\)-rays spectrum is emitted from the target. Which of the following wavelength is absent in the \(X\)-ray spectrum, if the X-ray tube is operating at \(40000 \mathrm{~V}\) ?
1 \(0.5 \AA\)
2 \(1.0 \AA\)
3 \(0.25 \AA\)
4 \(1.5 \AA\)
Explanation:
C X-ray tube operating (V) \(=40000 \mathrm{~V}\) We know, \(\lambda_{\text {sutoff }} =\frac{\mathrm{hc}}{\mathrm{eV}}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 40000}\) \(\lambda_{\text {sutofi }}=0.31 \AA\) \(\lambda>0.31 \AA\) wavelength is absent in the \(\mathrm{X}\)-ray spectrum.
142619
An element with atomic number \(Z=11\) emits \(\mathrm{E}_\alpha\) - X-ray of wavelength \(\lambda\). The atomic number which emits \(\mathrm{E}_\alpha-\mathrm{X}\)-ray of wavelength \(4 \lambda\) is
1 4
2 6
3 11
4 44
Explanation:
B Given, Atomic \(\operatorname{Number}\left(Z_1\right)=1\) Wavelength \(\left(\lambda_1\right)=\lambda\) Wavelength of second element \(\left(\lambda_2\right)=4 \lambda\). Then atomic number of second element \(\left(Z_2\right)=\) ? From Moseley's law- \(\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})\) \(\mathrm{f}=\mathrm{a}^2(Z-b)^2\) \(\frac{\lambda_1}{\lambda_2}=\frac{\left(Z_2-1\right)}{\left(Z_1-1\right)} \quad\left(\because \mathrm{f}=\frac{1}{\lambda}\right)\) \(\frac{\lambda}{4 \lambda}=\frac{\left(Z_2-1\right)^2}{(11-1)^2}\) \(\frac{100}{4}=\left(Z_2-1\right)^2\) \(\mathrm{Z}_2-1=5\) \(Z_2=6\)
VITEEE-2013
Dual nature of radiation and Matter
142621
The graph between the square root of the frequency of a specific line of characteristic spectrum of \(\mathrm{X}\) - ray and the atomic number of the target will be
1
2
3
4
Explanation:
B Relationship between frequency of characteristic spectrum of X-ray and atomic number - \(v=\mathrm{a}^2(\mathrm{Z}-\mathrm{b})^2\) Where, \(v=\) frequency \(\mathrm{Z}=\) atomic number a \& b are constant \(\sqrt{v}=a Z-a b\) We know, straight line equation \(\mathrm{Y}=\mathrm{mx}+\mathrm{c}\) Comparing equation (i) \& (ii), we get graph like this
VITEEE-2012
Dual nature of radiation and Matter
142625
If an X-ray tube is operated at \(15 \mathrm{kV}\), then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be
142626
When a beam of accelerated electrons hits a target, a continuous \(\mathrm{X}\)-rays spectrum is emitted from the target. Which of the following wavelength is absent in the \(X\)-ray spectrum, if the X-ray tube is operating at \(40000 \mathrm{~V}\) ?
1 \(0.5 \AA\)
2 \(1.0 \AA\)
3 \(0.25 \AA\)
4 \(1.5 \AA\)
Explanation:
C X-ray tube operating (V) \(=40000 \mathrm{~V}\) We know, \(\lambda_{\text {sutoff }} =\frac{\mathrm{hc}}{\mathrm{eV}}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 40000}\) \(\lambda_{\text {sutofi }}=0.31 \AA\) \(\lambda>0.31 \AA\) wavelength is absent in the \(\mathrm{X}\)-ray spectrum.
142619
An element with atomic number \(Z=11\) emits \(\mathrm{E}_\alpha\) - X-ray of wavelength \(\lambda\). The atomic number which emits \(\mathrm{E}_\alpha-\mathrm{X}\)-ray of wavelength \(4 \lambda\) is
1 4
2 6
3 11
4 44
Explanation:
B Given, Atomic \(\operatorname{Number}\left(Z_1\right)=1\) Wavelength \(\left(\lambda_1\right)=\lambda\) Wavelength of second element \(\left(\lambda_2\right)=4 \lambda\). Then atomic number of second element \(\left(Z_2\right)=\) ? From Moseley's law- \(\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{Z}-\mathrm{b})\) \(\mathrm{f}=\mathrm{a}^2(Z-b)^2\) \(\frac{\lambda_1}{\lambda_2}=\frac{\left(Z_2-1\right)}{\left(Z_1-1\right)} \quad\left(\because \mathrm{f}=\frac{1}{\lambda}\right)\) \(\frac{\lambda}{4 \lambda}=\frac{\left(Z_2-1\right)^2}{(11-1)^2}\) \(\frac{100}{4}=\left(Z_2-1\right)^2\) \(\mathrm{Z}_2-1=5\) \(Z_2=6\)
VITEEE-2013
Dual nature of radiation and Matter
142621
The graph between the square root of the frequency of a specific line of characteristic spectrum of \(\mathrm{X}\) - ray and the atomic number of the target will be
1
2
3
4
Explanation:
B Relationship between frequency of characteristic spectrum of X-ray and atomic number - \(v=\mathrm{a}^2(\mathrm{Z}-\mathrm{b})^2\) Where, \(v=\) frequency \(\mathrm{Z}=\) atomic number a \& b are constant \(\sqrt{v}=a Z-a b\) We know, straight line equation \(\mathrm{Y}=\mathrm{mx}+\mathrm{c}\) Comparing equation (i) \& (ii), we get graph like this
VITEEE-2012
Dual nature of radiation and Matter
142625
If an X-ray tube is operated at \(15 \mathrm{kV}\), then the upper limit of the speed of the electron striking the target and lower limit of the X-ray produced will be
142626
When a beam of accelerated electrons hits a target, a continuous \(\mathrm{X}\)-rays spectrum is emitted from the target. Which of the following wavelength is absent in the \(X\)-ray spectrum, if the X-ray tube is operating at \(40000 \mathrm{~V}\) ?
1 \(0.5 \AA\)
2 \(1.0 \AA\)
3 \(0.25 \AA\)
4 \(1.5 \AA\)
Explanation:
C X-ray tube operating (V) \(=40000 \mathrm{~V}\) We know, \(\lambda_{\text {sutoff }} =\frac{\mathrm{hc}}{\mathrm{eV}}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19} \times 40000}\) \(\lambda_{\text {sutofi }}=0.31 \AA\) \(\lambda>0.31 \AA\) wavelength is absent in the \(\mathrm{X}\)-ray spectrum.
