239097
The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
1 2
2 4
3 6
4 8
Explanation:
: When $\mathrm{n}=2$ and $l=1$ then subshell is $2 \mathrm{p}$ the number of orbital's in p-subshell. $\begin{gathered} \Rightarrow(2 l+1)=(2 \times 1+1) \\ =3 \end{gathered}$ Total (maximum) number of electrons $\begin{aligned} & 2 \times \text { number of orbital's } \\ \Rightarrow & 2 \times 3 \\ \Rightarrow & 6 \end{aligned}$ (as each orbital contains 2 electrons)
NEET-1990
Structure of Atom
239113
Which of the following electron has minimum energy?
: (i) Energy of electron depends on sum of $\mathrm{n}+$ 1 values. (ii) For the same sum of $n+1$, the orbital with lower value of $\mathrm{n}$ is filled first. Lower be the sum, lower the energy of electron. (iii) Electrons are filled in increasing order of energy. Thus among the given options, $\mathrm{n}+1$ for : (a) $4+0=4$ (b) $5+0=5$ (c) $4+1=5$ (d) $3+2=5$ $4+0=4$ is the
UPTU/UPSEE-2017
Structure of Atom
239121
If radius of second Bohr orbit of the $\mathrm{He}^{+}$ion is $105.8 \mathrm{pm}$, what is the radius of third Bohr orbit of $\mathbf{L i}^{2+}$ ion?
239122
If the radius of the $3^{\text {rd }}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_4$. Then:
1 $r_4=\frac{9}{16} r_3$
2 $r_4=\frac{16}{9} r_3$
3 $r_4=\frac{3}{4} r_3$
4 $\mathrm{r}_4=\frac{4}{3} \mathrm{r}_3$
Explanation:
$: r=\frac{\mathrm{n}^2}{\mathrm{z}}$ For $\mathrm{H}-$ atom, we know $=1$ So, $\quad \mathrm{r} \propto \mathrm{n}^2$ $\mathrm{r}_3 \propto 9$ $r_4 \propto 16$ From ratio $=\frac{r_3}{r_4}=\frac{9}{16}$ $r_4=\frac{16}{9} r_3$
239097
The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
1 2
2 4
3 6
4 8
Explanation:
: When $\mathrm{n}=2$ and $l=1$ then subshell is $2 \mathrm{p}$ the number of orbital's in p-subshell. $\begin{gathered} \Rightarrow(2 l+1)=(2 \times 1+1) \\ =3 \end{gathered}$ Total (maximum) number of electrons $\begin{aligned} & 2 \times \text { number of orbital's } \\ \Rightarrow & 2 \times 3 \\ \Rightarrow & 6 \end{aligned}$ (as each orbital contains 2 electrons)
NEET-1990
Structure of Atom
239113
Which of the following electron has minimum energy?
: (i) Energy of electron depends on sum of $\mathrm{n}+$ 1 values. (ii) For the same sum of $n+1$, the orbital with lower value of $\mathrm{n}$ is filled first. Lower be the sum, lower the energy of electron. (iii) Electrons are filled in increasing order of energy. Thus among the given options, $\mathrm{n}+1$ for : (a) $4+0=4$ (b) $5+0=5$ (c) $4+1=5$ (d) $3+2=5$ $4+0=4$ is the
UPTU/UPSEE-2017
Structure of Atom
239121
If radius of second Bohr orbit of the $\mathrm{He}^{+}$ion is $105.8 \mathrm{pm}$, what is the radius of third Bohr orbit of $\mathbf{L i}^{2+}$ ion?
239122
If the radius of the $3^{\text {rd }}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_4$. Then:
1 $r_4=\frac{9}{16} r_3$
2 $r_4=\frac{16}{9} r_3$
3 $r_4=\frac{3}{4} r_3$
4 $\mathrm{r}_4=\frac{4}{3} \mathrm{r}_3$
Explanation:
$: r=\frac{\mathrm{n}^2}{\mathrm{z}}$ For $\mathrm{H}-$ atom, we know $=1$ So, $\quad \mathrm{r} \propto \mathrm{n}^2$ $\mathrm{r}_3 \propto 9$ $r_4 \propto 16$ From ratio $=\frac{r_3}{r_4}=\frac{9}{16}$ $r_4=\frac{16}{9} r_3$
239097
The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
1 2
2 4
3 6
4 8
Explanation:
: When $\mathrm{n}=2$ and $l=1$ then subshell is $2 \mathrm{p}$ the number of orbital's in p-subshell. $\begin{gathered} \Rightarrow(2 l+1)=(2 \times 1+1) \\ =3 \end{gathered}$ Total (maximum) number of electrons $\begin{aligned} & 2 \times \text { number of orbital's } \\ \Rightarrow & 2 \times 3 \\ \Rightarrow & 6 \end{aligned}$ (as each orbital contains 2 electrons)
NEET-1990
Structure of Atom
239113
Which of the following electron has minimum energy?
: (i) Energy of electron depends on sum of $\mathrm{n}+$ 1 values. (ii) For the same sum of $n+1$, the orbital with lower value of $\mathrm{n}$ is filled first. Lower be the sum, lower the energy of electron. (iii) Electrons are filled in increasing order of energy. Thus among the given options, $\mathrm{n}+1$ for : (a) $4+0=4$ (b) $5+0=5$ (c) $4+1=5$ (d) $3+2=5$ $4+0=4$ is the
UPTU/UPSEE-2017
Structure of Atom
239121
If radius of second Bohr orbit of the $\mathrm{He}^{+}$ion is $105.8 \mathrm{pm}$, what is the radius of third Bohr orbit of $\mathbf{L i}^{2+}$ ion?
239122
If the radius of the $3^{\text {rd }}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_4$. Then:
1 $r_4=\frac{9}{16} r_3$
2 $r_4=\frac{16}{9} r_3$
3 $r_4=\frac{3}{4} r_3$
4 $\mathrm{r}_4=\frac{4}{3} \mathrm{r}_3$
Explanation:
$: r=\frac{\mathrm{n}^2}{\mathrm{z}}$ For $\mathrm{H}-$ atom, we know $=1$ So, $\quad \mathrm{r} \propto \mathrm{n}^2$ $\mathrm{r}_3 \propto 9$ $r_4 \propto 16$ From ratio $=\frac{r_3}{r_4}=\frac{9}{16}$ $r_4=\frac{16}{9} r_3$
239097
The total number of electrons that can be accommodated in all the orbitals having principal quantum number 2 and azimuthal quantum number 1 are
1 2
2 4
3 6
4 8
Explanation:
: When $\mathrm{n}=2$ and $l=1$ then subshell is $2 \mathrm{p}$ the number of orbital's in p-subshell. $\begin{gathered} \Rightarrow(2 l+1)=(2 \times 1+1) \\ =3 \end{gathered}$ Total (maximum) number of electrons $\begin{aligned} & 2 \times \text { number of orbital's } \\ \Rightarrow & 2 \times 3 \\ \Rightarrow & 6 \end{aligned}$ (as each orbital contains 2 electrons)
NEET-1990
Structure of Atom
239113
Which of the following electron has minimum energy?
: (i) Energy of electron depends on sum of $\mathrm{n}+$ 1 values. (ii) For the same sum of $n+1$, the orbital with lower value of $\mathrm{n}$ is filled first. Lower be the sum, lower the energy of electron. (iii) Electrons are filled in increasing order of energy. Thus among the given options, $\mathrm{n}+1$ for : (a) $4+0=4$ (b) $5+0=5$ (c) $4+1=5$ (d) $3+2=5$ $4+0=4$ is the
UPTU/UPSEE-2017
Structure of Atom
239121
If radius of second Bohr orbit of the $\mathrm{He}^{+}$ion is $105.8 \mathrm{pm}$, what is the radius of third Bohr orbit of $\mathbf{L i}^{2+}$ ion?
239122
If the radius of the $3^{\text {rd }}$ Bohr's orbit of hydrogen atom is $r_3$ and the radius of $4^{\text {th }}$ Bohr's orbit is $r_4$. Then:
1 $r_4=\frac{9}{16} r_3$
2 $r_4=\frac{16}{9} r_3$
3 $r_4=\frac{3}{4} r_3$
4 $\mathrm{r}_4=\frac{4}{3} \mathrm{r}_3$
Explanation:
$: r=\frac{\mathrm{n}^2}{\mathrm{z}}$ For $\mathrm{H}-$ atom, we know $=1$ So, $\quad \mathrm{r} \propto \mathrm{n}^2$ $\mathrm{r}_3 \propto 9$ $r_4 \propto 16$ From ratio $=\frac{r_3}{r_4}=\frac{9}{16}$ $r_4=\frac{16}{9} r_3$
