2 $4.8 \times 10^{-19}$ stat coulomb $/ \mathrm{kg}$
3 $1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
4 $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
Explanation:
D Given, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ Specific charge $=\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$ Thomson obtained a value of $1.76 \mathrm{C} / \mathrm{kg}$ for the specific charge of the electron.
J and K CET- 2004
ATOMS
145299
The ionization energy of $\mathrm{Li}^{2+}$ is equal to
1 $9 \mathrm{hcR}$
2 $6 \mathrm{hcR}$
3 $2 \mathrm{hcR}$
4 hcR
Explanation:
A Given, $Z=3$ We know that, Ionization energy, $\mathrm{E}=\mathrm{RchZ}^{2}$ Where, $\mathrm{Z}=$ atomic number $\text { Where, } =\text { atomic number }$ $\mathrm{n} =\text { plank constant }$ $\mathrm{c} =\text { speed of light }$ $\mathrm{R} =\text { constant }$ $\therefore \quad \mathrm{E} =\mathrm{Rch}(3)^{2}$ $\mathrm{E} =9 \mathrm{hcR}$
JIPMER-2011
ATOMS
145300
A coin of mass $=3.11 \mathrm{~g}$ is made of pure copper whose molar mass is $63.5 \mathrm{~g} / \mathrm{mol}$. Total positive charge in the coin is $(Z=29)$ for copper and Avogadro's number is $\mathbf{6 . 0 2} \times 10^{23}$ atoms $/$ mole)
1 $137000 \mathrm{C}$
2 $157000 \mathrm{C}$
3 $197000 \mathrm{C}$
4 $127000 \mathrm{C}$
Explanation:
A In $63.5 \mathrm{~g}$, number of atoms $=6.02 \times 10^{23}$ In $3.11 \mathrm{~g}$, number of atoms $=\frac{6.02 \times 10^{23} \times 3.11}{63.5}$ Since, $Z=29$ Number of electrons $=29$ in each atom. $\therefore$ Number of electrons or protons in copper coin $=29 \times$ number of atoms $\therefore$ Charge in the coin $=29 \times($ no. of atoms $) \times 1.6 \times 10^{-19}$ C $\therefore \text { Charge } =\frac{29 \times 6.02 \times 10^{23} \times 3.11 \times 1.6 \times 10^{-19}}{63.5} \mathrm{C}$ $=13.7 \times 10^{4} \mathrm{C}$ $=137000 \mathrm{C}$
AMU-2003
ATOMS
145239
If $v_{n}$ and $v_{p}$ are orbital velocities in $n^{\text {th }}$ and $p^{\text {th }}$ orbit respectively, then the ratio $v_{p}: v_{n}$ is
1 $\frac{n^{2}}{p^{2}}$
2 $\frac{\mathrm{p}}{\mathrm{n}}$
3 $\frac{\mathrm{p}^{2}}{\mathrm{n}^{2}}$
4 $\frac{\mathrm{n}}{\mathrm{p}}$
Explanation:
D We know that, Velocity, $v=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hn}}$ $\therefore \mathrm{v} \propto \frac{1}{\mathrm{n}}$ $\therefore \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{p}}$
MHT-CET } 2020
ATOMS
145250
For a light nuclei, which of the following relation between the atomic number $(Z)$ and mass number $(A)$ is valid?
1 $\mathrm{A}=\mathrm{Z} / 2$
2 $\mathrm{Z}=\mathrm{A}$
3 $\mathrm{Z}=\mathrm{A} / 2$
4 $Z=A^{2}$
5 $A=Z^{2}$
Explanation:
C Given, Atomic number $=\mathrm{Z}$ Mass number $=\mathrm{A}$ For lighter nuclei, the relation between atomic number and mass number is - $\mathrm{Z}=\mathrm{A} / 2$
2 $4.8 \times 10^{-19}$ stat coulomb $/ \mathrm{kg}$
3 $1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
4 $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
Explanation:
D Given, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ Specific charge $=\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$ Thomson obtained a value of $1.76 \mathrm{C} / \mathrm{kg}$ for the specific charge of the electron.
J and K CET- 2004
ATOMS
145299
The ionization energy of $\mathrm{Li}^{2+}$ is equal to
1 $9 \mathrm{hcR}$
2 $6 \mathrm{hcR}$
3 $2 \mathrm{hcR}$
4 hcR
Explanation:
A Given, $Z=3$ We know that, Ionization energy, $\mathrm{E}=\mathrm{RchZ}^{2}$ Where, $\mathrm{Z}=$ atomic number $\text { Where, } =\text { atomic number }$ $\mathrm{n} =\text { plank constant }$ $\mathrm{c} =\text { speed of light }$ $\mathrm{R} =\text { constant }$ $\therefore \quad \mathrm{E} =\mathrm{Rch}(3)^{2}$ $\mathrm{E} =9 \mathrm{hcR}$
JIPMER-2011
ATOMS
145300
A coin of mass $=3.11 \mathrm{~g}$ is made of pure copper whose molar mass is $63.5 \mathrm{~g} / \mathrm{mol}$. Total positive charge in the coin is $(Z=29)$ for copper and Avogadro's number is $\mathbf{6 . 0 2} \times 10^{23}$ atoms $/$ mole)
1 $137000 \mathrm{C}$
2 $157000 \mathrm{C}$
3 $197000 \mathrm{C}$
4 $127000 \mathrm{C}$
Explanation:
A In $63.5 \mathrm{~g}$, number of atoms $=6.02 \times 10^{23}$ In $3.11 \mathrm{~g}$, number of atoms $=\frac{6.02 \times 10^{23} \times 3.11}{63.5}$ Since, $Z=29$ Number of electrons $=29$ in each atom. $\therefore$ Number of electrons or protons in copper coin $=29 \times$ number of atoms $\therefore$ Charge in the coin $=29 \times($ no. of atoms $) \times 1.6 \times 10^{-19}$ C $\therefore \text { Charge } =\frac{29 \times 6.02 \times 10^{23} \times 3.11 \times 1.6 \times 10^{-19}}{63.5} \mathrm{C}$ $=13.7 \times 10^{4} \mathrm{C}$ $=137000 \mathrm{C}$
AMU-2003
ATOMS
145239
If $v_{n}$ and $v_{p}$ are orbital velocities in $n^{\text {th }}$ and $p^{\text {th }}$ orbit respectively, then the ratio $v_{p}: v_{n}$ is
1 $\frac{n^{2}}{p^{2}}$
2 $\frac{\mathrm{p}}{\mathrm{n}}$
3 $\frac{\mathrm{p}^{2}}{\mathrm{n}^{2}}$
4 $\frac{\mathrm{n}}{\mathrm{p}}$
Explanation:
D We know that, Velocity, $v=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hn}}$ $\therefore \mathrm{v} \propto \frac{1}{\mathrm{n}}$ $\therefore \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{p}}$
MHT-CET } 2020
ATOMS
145250
For a light nuclei, which of the following relation between the atomic number $(Z)$ and mass number $(A)$ is valid?
1 $\mathrm{A}=\mathrm{Z} / 2$
2 $\mathrm{Z}=\mathrm{A}$
3 $\mathrm{Z}=\mathrm{A} / 2$
4 $Z=A^{2}$
5 $A=Z^{2}$
Explanation:
C Given, Atomic number $=\mathrm{Z}$ Mass number $=\mathrm{A}$ For lighter nuclei, the relation between atomic number and mass number is - $\mathrm{Z}=\mathrm{A} / 2$
2 $4.8 \times 10^{-19}$ stat coulomb $/ \mathrm{kg}$
3 $1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
4 $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
Explanation:
D Given, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ Specific charge $=\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$ Thomson obtained a value of $1.76 \mathrm{C} / \mathrm{kg}$ for the specific charge of the electron.
J and K CET- 2004
ATOMS
145299
The ionization energy of $\mathrm{Li}^{2+}$ is equal to
1 $9 \mathrm{hcR}$
2 $6 \mathrm{hcR}$
3 $2 \mathrm{hcR}$
4 hcR
Explanation:
A Given, $Z=3$ We know that, Ionization energy, $\mathrm{E}=\mathrm{RchZ}^{2}$ Where, $\mathrm{Z}=$ atomic number $\text { Where, } =\text { atomic number }$ $\mathrm{n} =\text { plank constant }$ $\mathrm{c} =\text { speed of light }$ $\mathrm{R} =\text { constant }$ $\therefore \quad \mathrm{E} =\mathrm{Rch}(3)^{2}$ $\mathrm{E} =9 \mathrm{hcR}$
JIPMER-2011
ATOMS
145300
A coin of mass $=3.11 \mathrm{~g}$ is made of pure copper whose molar mass is $63.5 \mathrm{~g} / \mathrm{mol}$. Total positive charge in the coin is $(Z=29)$ for copper and Avogadro's number is $\mathbf{6 . 0 2} \times 10^{23}$ atoms $/$ mole)
1 $137000 \mathrm{C}$
2 $157000 \mathrm{C}$
3 $197000 \mathrm{C}$
4 $127000 \mathrm{C}$
Explanation:
A In $63.5 \mathrm{~g}$, number of atoms $=6.02 \times 10^{23}$ In $3.11 \mathrm{~g}$, number of atoms $=\frac{6.02 \times 10^{23} \times 3.11}{63.5}$ Since, $Z=29$ Number of electrons $=29$ in each atom. $\therefore$ Number of electrons or protons in copper coin $=29 \times$ number of atoms $\therefore$ Charge in the coin $=29 \times($ no. of atoms $) \times 1.6 \times 10^{-19}$ C $\therefore \text { Charge } =\frac{29 \times 6.02 \times 10^{23} \times 3.11 \times 1.6 \times 10^{-19}}{63.5} \mathrm{C}$ $=13.7 \times 10^{4} \mathrm{C}$ $=137000 \mathrm{C}$
AMU-2003
ATOMS
145239
If $v_{n}$ and $v_{p}$ are orbital velocities in $n^{\text {th }}$ and $p^{\text {th }}$ orbit respectively, then the ratio $v_{p}: v_{n}$ is
1 $\frac{n^{2}}{p^{2}}$
2 $\frac{\mathrm{p}}{\mathrm{n}}$
3 $\frac{\mathrm{p}^{2}}{\mathrm{n}^{2}}$
4 $\frac{\mathrm{n}}{\mathrm{p}}$
Explanation:
D We know that, Velocity, $v=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hn}}$ $\therefore \mathrm{v} \propto \frac{1}{\mathrm{n}}$ $\therefore \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{p}}$
MHT-CET } 2020
ATOMS
145250
For a light nuclei, which of the following relation between the atomic number $(Z)$ and mass number $(A)$ is valid?
1 $\mathrm{A}=\mathrm{Z} / 2$
2 $\mathrm{Z}=\mathrm{A}$
3 $\mathrm{Z}=\mathrm{A} / 2$
4 $Z=A^{2}$
5 $A=Z^{2}$
Explanation:
C Given, Atomic number $=\mathrm{Z}$ Mass number $=\mathrm{A}$ For lighter nuclei, the relation between atomic number and mass number is - $\mathrm{Z}=\mathrm{A} / 2$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
ATOMS
145291
The specific charge of an electron is
1 $1.6 \times 10^{-19}$ coulomb $/ \mathrm{kg}$
2 $4.8 \times 10^{-19}$ stat coulomb $/ \mathrm{kg}$
3 $1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
4 $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
Explanation:
D Given, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ Specific charge $=\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$ Thomson obtained a value of $1.76 \mathrm{C} / \mathrm{kg}$ for the specific charge of the electron.
J and K CET- 2004
ATOMS
145299
The ionization energy of $\mathrm{Li}^{2+}$ is equal to
1 $9 \mathrm{hcR}$
2 $6 \mathrm{hcR}$
3 $2 \mathrm{hcR}$
4 hcR
Explanation:
A Given, $Z=3$ We know that, Ionization energy, $\mathrm{E}=\mathrm{RchZ}^{2}$ Where, $\mathrm{Z}=$ atomic number $\text { Where, } =\text { atomic number }$ $\mathrm{n} =\text { plank constant }$ $\mathrm{c} =\text { speed of light }$ $\mathrm{R} =\text { constant }$ $\therefore \quad \mathrm{E} =\mathrm{Rch}(3)^{2}$ $\mathrm{E} =9 \mathrm{hcR}$
JIPMER-2011
ATOMS
145300
A coin of mass $=3.11 \mathrm{~g}$ is made of pure copper whose molar mass is $63.5 \mathrm{~g} / \mathrm{mol}$. Total positive charge in the coin is $(Z=29)$ for copper and Avogadro's number is $\mathbf{6 . 0 2} \times 10^{23}$ atoms $/$ mole)
1 $137000 \mathrm{C}$
2 $157000 \mathrm{C}$
3 $197000 \mathrm{C}$
4 $127000 \mathrm{C}$
Explanation:
A In $63.5 \mathrm{~g}$, number of atoms $=6.02 \times 10^{23}$ In $3.11 \mathrm{~g}$, number of atoms $=\frac{6.02 \times 10^{23} \times 3.11}{63.5}$ Since, $Z=29$ Number of electrons $=29$ in each atom. $\therefore$ Number of electrons or protons in copper coin $=29 \times$ number of atoms $\therefore$ Charge in the coin $=29 \times($ no. of atoms $) \times 1.6 \times 10^{-19}$ C $\therefore \text { Charge } =\frac{29 \times 6.02 \times 10^{23} \times 3.11 \times 1.6 \times 10^{-19}}{63.5} \mathrm{C}$ $=13.7 \times 10^{4} \mathrm{C}$ $=137000 \mathrm{C}$
AMU-2003
ATOMS
145239
If $v_{n}$ and $v_{p}$ are orbital velocities in $n^{\text {th }}$ and $p^{\text {th }}$ orbit respectively, then the ratio $v_{p}: v_{n}$ is
1 $\frac{n^{2}}{p^{2}}$
2 $\frac{\mathrm{p}}{\mathrm{n}}$
3 $\frac{\mathrm{p}^{2}}{\mathrm{n}^{2}}$
4 $\frac{\mathrm{n}}{\mathrm{p}}$
Explanation:
D We know that, Velocity, $v=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hn}}$ $\therefore \mathrm{v} \propto \frac{1}{\mathrm{n}}$ $\therefore \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{p}}$
MHT-CET } 2020
ATOMS
145250
For a light nuclei, which of the following relation between the atomic number $(Z)$ and mass number $(A)$ is valid?
1 $\mathrm{A}=\mathrm{Z} / 2$
2 $\mathrm{Z}=\mathrm{A}$
3 $\mathrm{Z}=\mathrm{A} / 2$
4 $Z=A^{2}$
5 $A=Z^{2}$
Explanation:
C Given, Atomic number $=\mathrm{Z}$ Mass number $=\mathrm{A}$ For lighter nuclei, the relation between atomic number and mass number is - $\mathrm{Z}=\mathrm{A} / 2$
2 $4.8 \times 10^{-19}$ stat coulomb $/ \mathrm{kg}$
3 $1.76 \times 10^{-11}$ coulomb $/ \mathrm{kg}$
4 $1.76 \times 10^{11}$ coulomb $/ \mathrm{kg}$
Explanation:
D Given, $\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}$ $\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ Specific charge $=\frac{\mathrm{e}}{\mathrm{m}}=\frac{1.6 \times 10^{-19}}{9.1 \times 10^{-31}}$ $=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$ Thomson obtained a value of $1.76 \mathrm{C} / \mathrm{kg}$ for the specific charge of the electron.
J and K CET- 2004
ATOMS
145299
The ionization energy of $\mathrm{Li}^{2+}$ is equal to
1 $9 \mathrm{hcR}$
2 $6 \mathrm{hcR}$
3 $2 \mathrm{hcR}$
4 hcR
Explanation:
A Given, $Z=3$ We know that, Ionization energy, $\mathrm{E}=\mathrm{RchZ}^{2}$ Where, $\mathrm{Z}=$ atomic number $\text { Where, } =\text { atomic number }$ $\mathrm{n} =\text { plank constant }$ $\mathrm{c} =\text { speed of light }$ $\mathrm{R} =\text { constant }$ $\therefore \quad \mathrm{E} =\mathrm{Rch}(3)^{2}$ $\mathrm{E} =9 \mathrm{hcR}$
JIPMER-2011
ATOMS
145300
A coin of mass $=3.11 \mathrm{~g}$ is made of pure copper whose molar mass is $63.5 \mathrm{~g} / \mathrm{mol}$. Total positive charge in the coin is $(Z=29)$ for copper and Avogadro's number is $\mathbf{6 . 0 2} \times 10^{23}$ atoms $/$ mole)
1 $137000 \mathrm{C}$
2 $157000 \mathrm{C}$
3 $197000 \mathrm{C}$
4 $127000 \mathrm{C}$
Explanation:
A In $63.5 \mathrm{~g}$, number of atoms $=6.02 \times 10^{23}$ In $3.11 \mathrm{~g}$, number of atoms $=\frac{6.02 \times 10^{23} \times 3.11}{63.5}$ Since, $Z=29$ Number of electrons $=29$ in each atom. $\therefore$ Number of electrons or protons in copper coin $=29 \times$ number of atoms $\therefore$ Charge in the coin $=29 \times($ no. of atoms $) \times 1.6 \times 10^{-19}$ C $\therefore \text { Charge } =\frac{29 \times 6.02 \times 10^{23} \times 3.11 \times 1.6 \times 10^{-19}}{63.5} \mathrm{C}$ $=13.7 \times 10^{4} \mathrm{C}$ $=137000 \mathrm{C}$
AMU-2003
ATOMS
145239
If $v_{n}$ and $v_{p}$ are orbital velocities in $n^{\text {th }}$ and $p^{\text {th }}$ orbit respectively, then the ratio $v_{p}: v_{n}$ is
1 $\frac{n^{2}}{p^{2}}$
2 $\frac{\mathrm{p}}{\mathrm{n}}$
3 $\frac{\mathrm{p}^{2}}{\mathrm{n}^{2}}$
4 $\frac{\mathrm{n}}{\mathrm{p}}$
Explanation:
D We know that, Velocity, $v=\frac{\mathrm{e}^{2}}{2 \varepsilon_{0} \mathrm{hn}}$ $\therefore \mathrm{v} \propto \frac{1}{\mathrm{n}}$ $\therefore \frac{\mathrm{v}_{\mathrm{p}}}{\mathrm{v}_{\mathrm{n}}}=\frac{\mathrm{n}}{\mathrm{p}}$
MHT-CET } 2020
ATOMS
145250
For a light nuclei, which of the following relation between the atomic number $(Z)$ and mass number $(A)$ is valid?
1 $\mathrm{A}=\mathrm{Z} / 2$
2 $\mathrm{Z}=\mathrm{A}$
3 $\mathrm{Z}=\mathrm{A} / 2$
4 $Z=A^{2}$
5 $A=Z^{2}$
Explanation:
C Given, Atomic number $=\mathrm{Z}$ Mass number $=\mathrm{A}$ For lighter nuclei, the relation between atomic number and mass number is - $\mathrm{Z}=\mathrm{A} / 2$