239124
The energy of an electrons in the $3^{\text {rd }}$ orbit of an atom is $-\mathbf{E}$. The energy of an electron in the first orbit will be
1 $-3 \mathrm{E}$
2 $-\frac{E}{3}$
3 $-\frac{E}{9}$
4 $-9 \mathrm{E}$
Explanation:
Given that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6 \mathrm{z}^2}{\mathrm{n}^2} \mathrm{eV} \text { atom }^{-1}$ For $3^{\text {rd }}$ Oribit $\begin{aligned} & \mathrm{E}_3=-\mathrm{E}=\frac{-13.6(z)^2}{3^2} \\ & \mathrm{E}_1=\frac{-13.6(z)^2}{(1)^2} \end{aligned}$ Divided equation (i) and (ii) we get $\begin{aligned} & \frac{-E}{E_1}=\frac{-13.6}{3^2} \times \frac{1^2}{-13.6} \\ & \mathrm{E}_1=-9 \mathrm{E} \end{aligned}$
MPPET- 2009
Structure of Atom
239125
If the radius of $1 s$ electron orbit of a hydrogen atom is $53 \mathrm{pm}$, then the radius of the $3 \mathrm{p}$ electron orbit would be
1 $0.477 \mathrm{~nm}$
2 $477 \mathrm{~nm}$
3 $159 \mathrm{pm}$
4 $17.66 \mathrm{~nm}$
Explanation:
Given that, $\mathrm{n}=1, \mathrm{r}_1=53 \mathrm{pm}, \mathrm{n}=3, \mathrm{r}_3=$ ? Now, from the radii expression- $r_n=\frac{52.9 n^2}{\mathrm{z}} \mathrm{pm}$ Where, $\mathrm{n}=$ no. of orbit. $\mathrm{z}=$ atomic number $\therefore \quad \mathrm{r}_3=\frac{52.9 \times 9}{1} \mathrm{pm}$ $\mathrm{r}_3=476.1 \mathrm{pm}$ or $\quad r_3=0.476 \mathrm{~nm}$
SCRA-2014
Structure of Atom
239126
The velocity of an electron in the first Bohr orbit is $v_1$. What is its velocity in the third Bohr's orbit?
1 $v_1 / 9$
2 $\mathrm{v}_1 / 3$
3 $\mathrm{v}_1$
4 $3 \mathrm{v}_1$
Explanation:
The velocity of an electron in the $\mathrm{n}^{\text {th }}$ stationary orbit for $\mathrm{H}$ - like species is expressed as, $\mathrm{V}_{\mathrm{n}}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} \mathrm{sec}^{-1}$ Here, $\quad z=1$ and $n=3$ So, Velocity in the third $\left\{\because\right.$ velocity of $\mathrm{e}^{-}$in first Bohr orbit $=\mathrm{V}_1$ \} Bohr's orbit $=\frac{\mathrm{v}_1}{3}$
SCRA-2010
Structure of Atom
239128
The basic assumption of Bohr's model of hydrogen atom is that
1 the energy of the electron is quantised
2 the angular momentum of the electron is quantised
3 the radial distance of the electron is quantised
4 the orbital velocity of the electron is quantized
Explanation:
Bohr's Model- (1) Electrons are revolving around nucleus $\mathrm{K}, \mathrm{L}, \mathrm{M}, \mathrm{N}$ in specific energy level with centripetal force given by nucleus. $\frac{\mathrm{mu}^2}{\mathrm{r}}=\frac{\mathrm{ZKe}^2}{\mathrm{r}^2}$ (2) Only those orbits are allowed whose angular momentum (p) is integral multiple of $\frac{\mathrm{h}}{2 \pi}$. $\mathrm{p}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
239124
The energy of an electrons in the $3^{\text {rd }}$ orbit of an atom is $-\mathbf{E}$. The energy of an electron in the first orbit will be
1 $-3 \mathrm{E}$
2 $-\frac{E}{3}$
3 $-\frac{E}{9}$
4 $-9 \mathrm{E}$
Explanation:
Given that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6 \mathrm{z}^2}{\mathrm{n}^2} \mathrm{eV} \text { atom }^{-1}$ For $3^{\text {rd }}$ Oribit $\begin{aligned} & \mathrm{E}_3=-\mathrm{E}=\frac{-13.6(z)^2}{3^2} \\ & \mathrm{E}_1=\frac{-13.6(z)^2}{(1)^2} \end{aligned}$ Divided equation (i) and (ii) we get $\begin{aligned} & \frac{-E}{E_1}=\frac{-13.6}{3^2} \times \frac{1^2}{-13.6} \\ & \mathrm{E}_1=-9 \mathrm{E} \end{aligned}$
MPPET- 2009
Structure of Atom
239125
If the radius of $1 s$ electron orbit of a hydrogen atom is $53 \mathrm{pm}$, then the radius of the $3 \mathrm{p}$ electron orbit would be
1 $0.477 \mathrm{~nm}$
2 $477 \mathrm{~nm}$
3 $159 \mathrm{pm}$
4 $17.66 \mathrm{~nm}$
Explanation:
Given that, $\mathrm{n}=1, \mathrm{r}_1=53 \mathrm{pm}, \mathrm{n}=3, \mathrm{r}_3=$ ? Now, from the radii expression- $r_n=\frac{52.9 n^2}{\mathrm{z}} \mathrm{pm}$ Where, $\mathrm{n}=$ no. of orbit. $\mathrm{z}=$ atomic number $\therefore \quad \mathrm{r}_3=\frac{52.9 \times 9}{1} \mathrm{pm}$ $\mathrm{r}_3=476.1 \mathrm{pm}$ or $\quad r_3=0.476 \mathrm{~nm}$
SCRA-2014
Structure of Atom
239126
The velocity of an electron in the first Bohr orbit is $v_1$. What is its velocity in the third Bohr's orbit?
1 $v_1 / 9$
2 $\mathrm{v}_1 / 3$
3 $\mathrm{v}_1$
4 $3 \mathrm{v}_1$
Explanation:
The velocity of an electron in the $\mathrm{n}^{\text {th }}$ stationary orbit for $\mathrm{H}$ - like species is expressed as, $\mathrm{V}_{\mathrm{n}}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} \mathrm{sec}^{-1}$ Here, $\quad z=1$ and $n=3$ So, Velocity in the third $\left\{\because\right.$ velocity of $\mathrm{e}^{-}$in first Bohr orbit $=\mathrm{V}_1$ \} Bohr's orbit $=\frac{\mathrm{v}_1}{3}$
SCRA-2010
Structure of Atom
239128
The basic assumption of Bohr's model of hydrogen atom is that
1 the energy of the electron is quantised
2 the angular momentum of the electron is quantised
3 the radial distance of the electron is quantised
4 the orbital velocity of the electron is quantized
Explanation:
Bohr's Model- (1) Electrons are revolving around nucleus $\mathrm{K}, \mathrm{L}, \mathrm{M}, \mathrm{N}$ in specific energy level with centripetal force given by nucleus. $\frac{\mathrm{mu}^2}{\mathrm{r}}=\frac{\mathrm{ZKe}^2}{\mathrm{r}^2}$ (2) Only those orbits are allowed whose angular momentum (p) is integral multiple of $\frac{\mathrm{h}}{2 \pi}$. $\mathrm{p}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
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Structure of Atom
239124
The energy of an electrons in the $3^{\text {rd }}$ orbit of an atom is $-\mathbf{E}$. The energy of an electron in the first orbit will be
1 $-3 \mathrm{E}$
2 $-\frac{E}{3}$
3 $-\frac{E}{9}$
4 $-9 \mathrm{E}$
Explanation:
Given that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6 \mathrm{z}^2}{\mathrm{n}^2} \mathrm{eV} \text { atom }^{-1}$ For $3^{\text {rd }}$ Oribit $\begin{aligned} & \mathrm{E}_3=-\mathrm{E}=\frac{-13.6(z)^2}{3^2} \\ & \mathrm{E}_1=\frac{-13.6(z)^2}{(1)^2} \end{aligned}$ Divided equation (i) and (ii) we get $\begin{aligned} & \frac{-E}{E_1}=\frac{-13.6}{3^2} \times \frac{1^2}{-13.6} \\ & \mathrm{E}_1=-9 \mathrm{E} \end{aligned}$
MPPET- 2009
Structure of Atom
239125
If the radius of $1 s$ electron orbit of a hydrogen atom is $53 \mathrm{pm}$, then the radius of the $3 \mathrm{p}$ electron orbit would be
1 $0.477 \mathrm{~nm}$
2 $477 \mathrm{~nm}$
3 $159 \mathrm{pm}$
4 $17.66 \mathrm{~nm}$
Explanation:
Given that, $\mathrm{n}=1, \mathrm{r}_1=53 \mathrm{pm}, \mathrm{n}=3, \mathrm{r}_3=$ ? Now, from the radii expression- $r_n=\frac{52.9 n^2}{\mathrm{z}} \mathrm{pm}$ Where, $\mathrm{n}=$ no. of orbit. $\mathrm{z}=$ atomic number $\therefore \quad \mathrm{r}_3=\frac{52.9 \times 9}{1} \mathrm{pm}$ $\mathrm{r}_3=476.1 \mathrm{pm}$ or $\quad r_3=0.476 \mathrm{~nm}$
SCRA-2014
Structure of Atom
239126
The velocity of an electron in the first Bohr orbit is $v_1$. What is its velocity in the third Bohr's orbit?
1 $v_1 / 9$
2 $\mathrm{v}_1 / 3$
3 $\mathrm{v}_1$
4 $3 \mathrm{v}_1$
Explanation:
The velocity of an electron in the $\mathrm{n}^{\text {th }}$ stationary orbit for $\mathrm{H}$ - like species is expressed as, $\mathrm{V}_{\mathrm{n}}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} \mathrm{sec}^{-1}$ Here, $\quad z=1$ and $n=3$ So, Velocity in the third $\left\{\because\right.$ velocity of $\mathrm{e}^{-}$in first Bohr orbit $=\mathrm{V}_1$ \} Bohr's orbit $=\frac{\mathrm{v}_1}{3}$
SCRA-2010
Structure of Atom
239128
The basic assumption of Bohr's model of hydrogen atom is that
1 the energy of the electron is quantised
2 the angular momentum of the electron is quantised
3 the radial distance of the electron is quantised
4 the orbital velocity of the electron is quantized
Explanation:
Bohr's Model- (1) Electrons are revolving around nucleus $\mathrm{K}, \mathrm{L}, \mathrm{M}, \mathrm{N}$ in specific energy level with centripetal force given by nucleus. $\frac{\mathrm{mu}^2}{\mathrm{r}}=\frac{\mathrm{ZKe}^2}{\mathrm{r}^2}$ (2) Only those orbits are allowed whose angular momentum (p) is integral multiple of $\frac{\mathrm{h}}{2 \pi}$. $\mathrm{p}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$
239124
The energy of an electrons in the $3^{\text {rd }}$ orbit of an atom is $-\mathbf{E}$. The energy of an electron in the first orbit will be
1 $-3 \mathrm{E}$
2 $-\frac{E}{3}$
3 $-\frac{E}{9}$
4 $-9 \mathrm{E}$
Explanation:
Given that, $\mathrm{E}_{\mathrm{n}}=-\frac{13.6 \mathrm{z}^2}{\mathrm{n}^2} \mathrm{eV} \text { atom }^{-1}$ For $3^{\text {rd }}$ Oribit $\begin{aligned} & \mathrm{E}_3=-\mathrm{E}=\frac{-13.6(z)^2}{3^2} \\ & \mathrm{E}_1=\frac{-13.6(z)^2}{(1)^2} \end{aligned}$ Divided equation (i) and (ii) we get $\begin{aligned} & \frac{-E}{E_1}=\frac{-13.6}{3^2} \times \frac{1^2}{-13.6} \\ & \mathrm{E}_1=-9 \mathrm{E} \end{aligned}$
MPPET- 2009
Structure of Atom
239125
If the radius of $1 s$ electron orbit of a hydrogen atom is $53 \mathrm{pm}$, then the radius of the $3 \mathrm{p}$ electron orbit would be
1 $0.477 \mathrm{~nm}$
2 $477 \mathrm{~nm}$
3 $159 \mathrm{pm}$
4 $17.66 \mathrm{~nm}$
Explanation:
Given that, $\mathrm{n}=1, \mathrm{r}_1=53 \mathrm{pm}, \mathrm{n}=3, \mathrm{r}_3=$ ? Now, from the radii expression- $r_n=\frac{52.9 n^2}{\mathrm{z}} \mathrm{pm}$ Where, $\mathrm{n}=$ no. of orbit. $\mathrm{z}=$ atomic number $\therefore \quad \mathrm{r}_3=\frac{52.9 \times 9}{1} \mathrm{pm}$ $\mathrm{r}_3=476.1 \mathrm{pm}$ or $\quad r_3=0.476 \mathrm{~nm}$
SCRA-2014
Structure of Atom
239126
The velocity of an electron in the first Bohr orbit is $v_1$. What is its velocity in the third Bohr's orbit?
1 $v_1 / 9$
2 $\mathrm{v}_1 / 3$
3 $\mathrm{v}_1$
4 $3 \mathrm{v}_1$
Explanation:
The velocity of an electron in the $\mathrm{n}^{\text {th }}$ stationary orbit for $\mathrm{H}$ - like species is expressed as, $\mathrm{V}_{\mathrm{n}}=2.18 \times 10^6 \frac{\mathrm{Z}}{\mathrm{n}} \mathrm{m} \mathrm{sec}^{-1}$ Here, $\quad z=1$ and $n=3$ So, Velocity in the third $\left\{\because\right.$ velocity of $\mathrm{e}^{-}$in first Bohr orbit $=\mathrm{V}_1$ \} Bohr's orbit $=\frac{\mathrm{v}_1}{3}$
SCRA-2010
Structure of Atom
239128
The basic assumption of Bohr's model of hydrogen atom is that
1 the energy of the electron is quantised
2 the angular momentum of the electron is quantised
3 the radial distance of the electron is quantised
4 the orbital velocity of the electron is quantized
Explanation:
Bohr's Model- (1) Electrons are revolving around nucleus $\mathrm{K}, \mathrm{L}, \mathrm{M}, \mathrm{N}$ in specific energy level with centripetal force given by nucleus. $\frac{\mathrm{mu}^2}{\mathrm{r}}=\frac{\mathrm{ZKe}^2}{\mathrm{r}^2}$ (2) Only those orbits are allowed whose angular momentum (p) is integral multiple of $\frac{\mathrm{h}}{2 \pi}$. $\mathrm{p}=\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$