145246
The frequency of $K_{a}$ line of a source of atomic number $z$ is proportional to
1 $\mathrm{z}^{2}$
2 $(z-1)^{2}$
3 $\frac{1}{\mathrm{z}}$
4 $\mathrm{z}$
Explanation:
B According to Moseley's law - $\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{z}-\mathrm{b})$ Where, $\mathrm{z}=$ atomic number $f=\text { frequency }$ $f \propto(z-b)^{2}$ $b=1$ For $\quad b=1$ So, $\quad f \propto(z-1)^{2}$
HP CET-2018
ATOMS
145247
If the kinetic energy of an electron of mass 9.0 $\times 10^{-31} \mathrm{~kg}$ is $8.0 \times 10^{-25} \mathrm{~J}$, the wavelength of this electron in $\mathrm{nm}$ is
A Given, Radius of $\mathrm{A} l$ nucleus $=\mathrm{R}_{1}$ Radius of ${ }_{33} \mathrm{Te}^{125}$ nucleus $=\mathrm{R}_{2}$ Mass of $\mathrm{A} l\left(\mathrm{~A}_{1}\right)=27$ Mass of ${ }_{33} \mathrm{Te}^{125}\left(\mathrm{~A}_{2}\right)=125$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\because \quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{125}\right)^{1 / 3} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ $\mathrm{R}_{2}=\frac{5}{3} \mathrm{R}_{1}$
BCECE-2018
ATOMS
145249
If elements with principal quantum number $n$ $>4$ were not allowed in nature, the number of possible elements would be
1 60
2 32
3 4
4 64
Explanation:
A The maximum number of electron in an orbit are $\mathrm{N}=2(\mathrm{n})^{2}$ Where, $\mathrm{N}=$ number of electrons in a particular orbit $\mathrm{n}=$ quantum number of the orbit For first orbit, number of electrons will be $\mathrm{N}=2(\mathrm{n})^{2}=2(1)^{2}=2$ For second, third and fourth orbit- $\mathrm{N}=2(2)^{2}=8$ $\mathrm{~N}=2(3)^{2}=18$ $\mathrm{~N}=2(4)^{2}=32$ The total number of elements is given by the total number of electrons. $\mathrm{N}=2+8+18+32=60$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145246
The frequency of $K_{a}$ line of a source of atomic number $z$ is proportional to
1 $\mathrm{z}^{2}$
2 $(z-1)^{2}$
3 $\frac{1}{\mathrm{z}}$
4 $\mathrm{z}$
Explanation:
B According to Moseley's law - $\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{z}-\mathrm{b})$ Where, $\mathrm{z}=$ atomic number $f=\text { frequency }$ $f \propto(z-b)^{2}$ $b=1$ For $\quad b=1$ So, $\quad f \propto(z-1)^{2}$
HP CET-2018
ATOMS
145247
If the kinetic energy of an electron of mass 9.0 $\times 10^{-31} \mathrm{~kg}$ is $8.0 \times 10^{-25} \mathrm{~J}$, the wavelength of this electron in $\mathrm{nm}$ is
A Given, Radius of $\mathrm{A} l$ nucleus $=\mathrm{R}_{1}$ Radius of ${ }_{33} \mathrm{Te}^{125}$ nucleus $=\mathrm{R}_{2}$ Mass of $\mathrm{A} l\left(\mathrm{~A}_{1}\right)=27$ Mass of ${ }_{33} \mathrm{Te}^{125}\left(\mathrm{~A}_{2}\right)=125$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\because \quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{125}\right)^{1 / 3} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ $\mathrm{R}_{2}=\frac{5}{3} \mathrm{R}_{1}$
BCECE-2018
ATOMS
145249
If elements with principal quantum number $n$ $>4$ were not allowed in nature, the number of possible elements would be
1 60
2 32
3 4
4 64
Explanation:
A The maximum number of electron in an orbit are $\mathrm{N}=2(\mathrm{n})^{2}$ Where, $\mathrm{N}=$ number of electrons in a particular orbit $\mathrm{n}=$ quantum number of the orbit For first orbit, number of electrons will be $\mathrm{N}=2(\mathrm{n})^{2}=2(1)^{2}=2$ For second, third and fourth orbit- $\mathrm{N}=2(2)^{2}=8$ $\mathrm{~N}=2(3)^{2}=18$ $\mathrm{~N}=2(4)^{2}=32$ The total number of elements is given by the total number of electrons. $\mathrm{N}=2+8+18+32=60$
145246
The frequency of $K_{a}$ line of a source of atomic number $z$ is proportional to
1 $\mathrm{z}^{2}$
2 $(z-1)^{2}$
3 $\frac{1}{\mathrm{z}}$
4 $\mathrm{z}$
Explanation:
B According to Moseley's law - $\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{z}-\mathrm{b})$ Where, $\mathrm{z}=$ atomic number $f=\text { frequency }$ $f \propto(z-b)^{2}$ $b=1$ For $\quad b=1$ So, $\quad f \propto(z-1)^{2}$
HP CET-2018
ATOMS
145247
If the kinetic energy of an electron of mass 9.0 $\times 10^{-31} \mathrm{~kg}$ is $8.0 \times 10^{-25} \mathrm{~J}$, the wavelength of this electron in $\mathrm{nm}$ is
A Given, Radius of $\mathrm{A} l$ nucleus $=\mathrm{R}_{1}$ Radius of ${ }_{33} \mathrm{Te}^{125}$ nucleus $=\mathrm{R}_{2}$ Mass of $\mathrm{A} l\left(\mathrm{~A}_{1}\right)=27$ Mass of ${ }_{33} \mathrm{Te}^{125}\left(\mathrm{~A}_{2}\right)=125$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\because \quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{125}\right)^{1 / 3} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ $\mathrm{R}_{2}=\frac{5}{3} \mathrm{R}_{1}$
BCECE-2018
ATOMS
145249
If elements with principal quantum number $n$ $>4$ were not allowed in nature, the number of possible elements would be
1 60
2 32
3 4
4 64
Explanation:
A The maximum number of electron in an orbit are $\mathrm{N}=2(\mathrm{n})^{2}$ Where, $\mathrm{N}=$ number of electrons in a particular orbit $\mathrm{n}=$ quantum number of the orbit For first orbit, number of electrons will be $\mathrm{N}=2(\mathrm{n})^{2}=2(1)^{2}=2$ For second, third and fourth orbit- $\mathrm{N}=2(2)^{2}=8$ $\mathrm{~N}=2(3)^{2}=18$ $\mathrm{~N}=2(4)^{2}=32$ The total number of elements is given by the total number of electrons. $\mathrm{N}=2+8+18+32=60$
145246
The frequency of $K_{a}$ line of a source of atomic number $z$ is proportional to
1 $\mathrm{z}^{2}$
2 $(z-1)^{2}$
3 $\frac{1}{\mathrm{z}}$
4 $\mathrm{z}$
Explanation:
B According to Moseley's law - $\sqrt{\mathrm{f}}=\mathrm{a}(\mathrm{z}-\mathrm{b})$ Where, $\mathrm{z}=$ atomic number $f=\text { frequency }$ $f \propto(z-b)^{2}$ $b=1$ For $\quad b=1$ So, $\quad f \propto(z-1)^{2}$
HP CET-2018
ATOMS
145247
If the kinetic energy of an electron of mass 9.0 $\times 10^{-31} \mathrm{~kg}$ is $8.0 \times 10^{-25} \mathrm{~J}$, the wavelength of this electron in $\mathrm{nm}$ is
A Given, Radius of $\mathrm{A} l$ nucleus $=\mathrm{R}_{1}$ Radius of ${ }_{33} \mathrm{Te}^{125}$ nucleus $=\mathrm{R}_{2}$ Mass of $\mathrm{A} l\left(\mathrm{~A}_{1}\right)=27$ Mass of ${ }_{33} \mathrm{Te}^{125}\left(\mathrm{~A}_{2}\right)=125$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\because \quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{\mathrm{A}_{1}}{\mathrm{~A}_{2}}\right)^{1 / 3}$ $\frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\left(\frac{27}{125}\right)^{1 / 3} \Rightarrow \frac{\mathrm{R}_{1}}{\mathrm{R}_{2}}=\frac{3}{5}$ $\mathrm{R}_{2}=\frac{5}{3} \mathrm{R}_{1}$
BCECE-2018
ATOMS
145249
If elements with principal quantum number $n$ $>4$ were not allowed in nature, the number of possible elements would be
1 60
2 32
3 4
4 64
Explanation:
A The maximum number of electron in an orbit are $\mathrm{N}=2(\mathrm{n})^{2}$ Where, $\mathrm{N}=$ number of electrons in a particular orbit $\mathrm{n}=$ quantum number of the orbit For first orbit, number of electrons will be $\mathrm{N}=2(\mathrm{n})^{2}=2(1)^{2}=2$ For second, third and fourth orbit- $\mathrm{N}=2(2)^{2}=8$ $\mathrm{~N}=2(3)^{2}=18$ $\mathrm{~N}=2(4)^{2}=32$ The total number of elements is given by the total number of electrons. $\mathrm{N}=2+8+18+32=60$