145282
When a hydrogen atom is raised from ground energy level to excited energy level, then
1 Potential energy increases and kinetic energy decreases
2 Kinetic energy increases and potential energy decreases
3 Both $\mathrm{KE}$ and $\mathrm{PE}$ increase
4 Both KE and PE decrease
Explanation:
A When a hydrogen atom is raised from ground energy level to excited energy level, then potential energy increases and kinetic energy decreases. We know that- Kinetic energy $=\frac{\mathrm{KZe}^{2}}{2 \mathrm{r}}$ $\mathrm{K} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When an electron is excited to higher energy level then $r$ increases and then K.E. decreases. Potential energy $=\frac{-\mathrm{KZe}^{2}}{\mathrm{r}}$ $\mathrm{P} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When radius increases then potential energy is also increase due to negative sign.
VITEEE-2015
ATOMS
145283
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is $(14)^{1 / 3}$. The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
B Given, The ration of radii of nucleus that of helium nucleus, $\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{H}_{2}}}=(14)^{1 / 3}$, number of neutron $=30$ Mass number of ${ }_{2} \mathrm{He}^{4}\left(\mathrm{~A}_{1}\right)=4$ We know that, the radius of nucleus is given by the formula $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\therefore \quad \frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{\mathrm{A}_{1}}\right)^{1 / 3}$ $(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$ $\therefore$ Atomic number of nucleus $=\mathrm{A}-$ no. of neutron $=56-30$ $=26$
VITEEE-2012
ATOMS
145287
The longest wavelength that can be analysed by a sodium chloride crystal of spacing $d=2.82 \AA$ in the second order is
1 $2.82 \AA$
2 $5.64 \AA$
3 $8.46 \AA$
4 $11.28 \AA$
Explanation:
A Given, $\mathrm{d}=2.82 \AA$, second order $(\mathrm{n})=2$ According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ Where, $\lambda=$ wave length of the $\mathrm{X}$-ray $\mathrm{d}=$ distance of crystal layers $\theta=$ incident angle $\mathrm{n}=$ integer (order) $\lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}$ In the second order $(n=2)$, the maximum wavelength possible in when $\sin \theta$ is maximum, Maximum value of $\sin \theta=1$ $\lambda_{\text {max }}=\frac{2 \mathrm{~d}}{\mathrm{n}}$ $\lambda_{\text {max }}=\frac{2 \times 2.82}{2}$ $\lambda_{\text {max }}=2.82 \AA$ The longest wavelength that can be analysed $=2.82 \AA$
VITEEE-2006
ATOMS
145289
Two nucleons are at a separation of $1 \times 10^{-5} \mathrm{~m}$. The net force between them is $F_{1}$, if both are neutrons, $F_{2}$ if both are protons, and $F_{3}$ if one is a proton and the other is a neutron. Then :
1 $\mathrm{F}_{1}=\mathrm{F}_{2}>\mathrm{F}_{3}$
2 $\mathrm{F}_{1}=\mathrm{F}_{3}>\mathrm{F}_{2}$
3 $\mathrm{F}_{2}>\mathrm{F}_{1}>\mathrm{F}_{3}$
4 $\mathrm{F}_{1}=\mathrm{F}_{2}=\mathrm{F}_{3}$
Explanation:
B Given, distance between two nucleons $=1 \times$ $10^{-15} \mathrm{~m}$. Number forces are the strongest forces of attraction which hold together the nucleons (neutrons and protons) of an atom, inspite of stronger electrostatic forces of repulsion between protons. $\mathrm{F}_{1} \rightarrow$ Force between two neutrons, there is only nuclear force acting. $\mathrm{F}_{3} \rightarrow$ Force between neutron and 1 proton, nuclear force. $\mathrm{F}_{2} \rightarrow$ Force between protons, there is two acting force, one is nuclear and other is electrostatic. Electrostatic force is repulsive in nature. So, $\mathrm{F}_{2}=$ nuclear force - electrostatic force. Hence, $F_{1}=F_{3}>F_{2}$
145282
When a hydrogen atom is raised from ground energy level to excited energy level, then
1 Potential energy increases and kinetic energy decreases
2 Kinetic energy increases and potential energy decreases
3 Both $\mathrm{KE}$ and $\mathrm{PE}$ increase
4 Both KE and PE decrease
Explanation:
A When a hydrogen atom is raised from ground energy level to excited energy level, then potential energy increases and kinetic energy decreases. We know that- Kinetic energy $=\frac{\mathrm{KZe}^{2}}{2 \mathrm{r}}$ $\mathrm{K} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When an electron is excited to higher energy level then $r$ increases and then K.E. decreases. Potential energy $=\frac{-\mathrm{KZe}^{2}}{\mathrm{r}}$ $\mathrm{P} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When radius increases then potential energy is also increase due to negative sign.
VITEEE-2015
ATOMS
145283
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is $(14)^{1 / 3}$. The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
B Given, The ration of radii of nucleus that of helium nucleus, $\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{H}_{2}}}=(14)^{1 / 3}$, number of neutron $=30$ Mass number of ${ }_{2} \mathrm{He}^{4}\left(\mathrm{~A}_{1}\right)=4$ We know that, the radius of nucleus is given by the formula $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\therefore \quad \frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{\mathrm{A}_{1}}\right)^{1 / 3}$ $(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$ $\therefore$ Atomic number of nucleus $=\mathrm{A}-$ no. of neutron $=56-30$ $=26$
VITEEE-2012
ATOMS
145287
The longest wavelength that can be analysed by a sodium chloride crystal of spacing $d=2.82 \AA$ in the second order is
1 $2.82 \AA$
2 $5.64 \AA$
3 $8.46 \AA$
4 $11.28 \AA$
Explanation:
A Given, $\mathrm{d}=2.82 \AA$, second order $(\mathrm{n})=2$ According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ Where, $\lambda=$ wave length of the $\mathrm{X}$-ray $\mathrm{d}=$ distance of crystal layers $\theta=$ incident angle $\mathrm{n}=$ integer (order) $\lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}$ In the second order $(n=2)$, the maximum wavelength possible in when $\sin \theta$ is maximum, Maximum value of $\sin \theta=1$ $\lambda_{\text {max }}=\frac{2 \mathrm{~d}}{\mathrm{n}}$ $\lambda_{\text {max }}=\frac{2 \times 2.82}{2}$ $\lambda_{\text {max }}=2.82 \AA$ The longest wavelength that can be analysed $=2.82 \AA$
VITEEE-2006
ATOMS
145289
Two nucleons are at a separation of $1 \times 10^{-5} \mathrm{~m}$. The net force between them is $F_{1}$, if both are neutrons, $F_{2}$ if both are protons, and $F_{3}$ if one is a proton and the other is a neutron. Then :
1 $\mathrm{F}_{1}=\mathrm{F}_{2}>\mathrm{F}_{3}$
2 $\mathrm{F}_{1}=\mathrm{F}_{3}>\mathrm{F}_{2}$
3 $\mathrm{F}_{2}>\mathrm{F}_{1}>\mathrm{F}_{3}$
4 $\mathrm{F}_{1}=\mathrm{F}_{2}=\mathrm{F}_{3}$
Explanation:
B Given, distance between two nucleons $=1 \times$ $10^{-15} \mathrm{~m}$. Number forces are the strongest forces of attraction which hold together the nucleons (neutrons and protons) of an atom, inspite of stronger electrostatic forces of repulsion between protons. $\mathrm{F}_{1} \rightarrow$ Force between two neutrons, there is only nuclear force acting. $\mathrm{F}_{3} \rightarrow$ Force between neutron and 1 proton, nuclear force. $\mathrm{F}_{2} \rightarrow$ Force between protons, there is two acting force, one is nuclear and other is electrostatic. Electrostatic force is repulsive in nature. So, $\mathrm{F}_{2}=$ nuclear force - electrostatic force. Hence, $F_{1}=F_{3}>F_{2}$
145282
When a hydrogen atom is raised from ground energy level to excited energy level, then
1 Potential energy increases and kinetic energy decreases
2 Kinetic energy increases and potential energy decreases
3 Both $\mathrm{KE}$ and $\mathrm{PE}$ increase
4 Both KE and PE decrease
Explanation:
A When a hydrogen atom is raised from ground energy level to excited energy level, then potential energy increases and kinetic energy decreases. We know that- Kinetic energy $=\frac{\mathrm{KZe}^{2}}{2 \mathrm{r}}$ $\mathrm{K} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When an electron is excited to higher energy level then $r$ increases and then K.E. decreases. Potential energy $=\frac{-\mathrm{KZe}^{2}}{\mathrm{r}}$ $\mathrm{P} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When radius increases then potential energy is also increase due to negative sign.
VITEEE-2015
ATOMS
145283
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is $(14)^{1 / 3}$. The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
B Given, The ration of radii of nucleus that of helium nucleus, $\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{H}_{2}}}=(14)^{1 / 3}$, number of neutron $=30$ Mass number of ${ }_{2} \mathrm{He}^{4}\left(\mathrm{~A}_{1}\right)=4$ We know that, the radius of nucleus is given by the formula $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\therefore \quad \frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{\mathrm{A}_{1}}\right)^{1 / 3}$ $(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$ $\therefore$ Atomic number of nucleus $=\mathrm{A}-$ no. of neutron $=56-30$ $=26$
VITEEE-2012
ATOMS
145287
The longest wavelength that can be analysed by a sodium chloride crystal of spacing $d=2.82 \AA$ in the second order is
1 $2.82 \AA$
2 $5.64 \AA$
3 $8.46 \AA$
4 $11.28 \AA$
Explanation:
A Given, $\mathrm{d}=2.82 \AA$, second order $(\mathrm{n})=2$ According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ Where, $\lambda=$ wave length of the $\mathrm{X}$-ray $\mathrm{d}=$ distance of crystal layers $\theta=$ incident angle $\mathrm{n}=$ integer (order) $\lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}$ In the second order $(n=2)$, the maximum wavelength possible in when $\sin \theta$ is maximum, Maximum value of $\sin \theta=1$ $\lambda_{\text {max }}=\frac{2 \mathrm{~d}}{\mathrm{n}}$ $\lambda_{\text {max }}=\frac{2 \times 2.82}{2}$ $\lambda_{\text {max }}=2.82 \AA$ The longest wavelength that can be analysed $=2.82 \AA$
VITEEE-2006
ATOMS
145289
Two nucleons are at a separation of $1 \times 10^{-5} \mathrm{~m}$. The net force between them is $F_{1}$, if both are neutrons, $F_{2}$ if both are protons, and $F_{3}$ if one is a proton and the other is a neutron. Then :
1 $\mathrm{F}_{1}=\mathrm{F}_{2}>\mathrm{F}_{3}$
2 $\mathrm{F}_{1}=\mathrm{F}_{3}>\mathrm{F}_{2}$
3 $\mathrm{F}_{2}>\mathrm{F}_{1}>\mathrm{F}_{3}$
4 $\mathrm{F}_{1}=\mathrm{F}_{2}=\mathrm{F}_{3}$
Explanation:
B Given, distance between two nucleons $=1 \times$ $10^{-15} \mathrm{~m}$. Number forces are the strongest forces of attraction which hold together the nucleons (neutrons and protons) of an atom, inspite of stronger electrostatic forces of repulsion between protons. $\mathrm{F}_{1} \rightarrow$ Force between two neutrons, there is only nuclear force acting. $\mathrm{F}_{3} \rightarrow$ Force between neutron and 1 proton, nuclear force. $\mathrm{F}_{2} \rightarrow$ Force between protons, there is two acting force, one is nuclear and other is electrostatic. Electrostatic force is repulsive in nature. So, $\mathrm{F}_{2}=$ nuclear force - electrostatic force. Hence, $F_{1}=F_{3}>F_{2}$
145282
When a hydrogen atom is raised from ground energy level to excited energy level, then
1 Potential energy increases and kinetic energy decreases
2 Kinetic energy increases and potential energy decreases
3 Both $\mathrm{KE}$ and $\mathrm{PE}$ increase
4 Both KE and PE decrease
Explanation:
A When a hydrogen atom is raised from ground energy level to excited energy level, then potential energy increases and kinetic energy decreases. We know that- Kinetic energy $=\frac{\mathrm{KZe}^{2}}{2 \mathrm{r}}$ $\mathrm{K} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When an electron is excited to higher energy level then $r$ increases and then K.E. decreases. Potential energy $=\frac{-\mathrm{KZe}^{2}}{\mathrm{r}}$ $\mathrm{P} \cdot \mathrm{E} \cdot \propto \frac{1}{\mathrm{r}}$ When radius increases then potential energy is also increase due to negative sign.
VITEEE-2015
ATOMS
145283
Highly energetic electrons are bombarded on a target of an element containing 30 neutrons. The ratio of radii of nucleus to that of helium nucleus is $(14)^{1 / 3}$. The atomic number of nucleus will be
1 25
2 26
3 56
4 30
Explanation:
B Given, The ration of radii of nucleus that of helium nucleus, $\frac{\mathrm{R}}{\mathrm{R}_{\mathrm{H}_{2}}}=(14)^{1 / 3}$, number of neutron $=30$ Mass number of ${ }_{2} \mathrm{He}^{4}\left(\mathrm{~A}_{1}\right)=4$ We know that, the radius of nucleus is given by the formula $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant $\therefore \quad \frac{\mathrm{R}}{\mathrm{R}_{\mathrm{He}}}=\left(\frac{\mathrm{A}}{\mathrm{A}_{1}}\right)^{1 / 3}$ $(14)^{1 / 3}=\left(\frac{\mathrm{A}}{4}\right)^{1 / 3}$ $\therefore$ Atomic number of nucleus $=\mathrm{A}-$ no. of neutron $=56-30$ $=26$
VITEEE-2012
ATOMS
145287
The longest wavelength that can be analysed by a sodium chloride crystal of spacing $d=2.82 \AA$ in the second order is
1 $2.82 \AA$
2 $5.64 \AA$
3 $8.46 \AA$
4 $11.28 \AA$
Explanation:
A Given, $\mathrm{d}=2.82 \AA$, second order $(\mathrm{n})=2$ According to Bragg's law - $\mathrm{n} \lambda=2 \mathrm{~d} \sin \theta$ Where, $\lambda=$ wave length of the $\mathrm{X}$-ray $\mathrm{d}=$ distance of crystal layers $\theta=$ incident angle $\mathrm{n}=$ integer (order) $\lambda=\frac{2 \mathrm{~d} \sin \theta}{\mathrm{n}}$ In the second order $(n=2)$, the maximum wavelength possible in when $\sin \theta$ is maximum, Maximum value of $\sin \theta=1$ $\lambda_{\text {max }}=\frac{2 \mathrm{~d}}{\mathrm{n}}$ $\lambda_{\text {max }}=\frac{2 \times 2.82}{2}$ $\lambda_{\text {max }}=2.82 \AA$ The longest wavelength that can be analysed $=2.82 \AA$
VITEEE-2006
ATOMS
145289
Two nucleons are at a separation of $1 \times 10^{-5} \mathrm{~m}$. The net force between them is $F_{1}$, if both are neutrons, $F_{2}$ if both are protons, and $F_{3}$ if one is a proton and the other is a neutron. Then :
1 $\mathrm{F}_{1}=\mathrm{F}_{2}>\mathrm{F}_{3}$
2 $\mathrm{F}_{1}=\mathrm{F}_{3}>\mathrm{F}_{2}$
3 $\mathrm{F}_{2}>\mathrm{F}_{1}>\mathrm{F}_{3}$
4 $\mathrm{F}_{1}=\mathrm{F}_{2}=\mathrm{F}_{3}$
Explanation:
B Given, distance between two nucleons $=1 \times$ $10^{-15} \mathrm{~m}$. Number forces are the strongest forces of attraction which hold together the nucleons (neutrons and protons) of an atom, inspite of stronger electrostatic forces of repulsion between protons. $\mathrm{F}_{1} \rightarrow$ Force between two neutrons, there is only nuclear force acting. $\mathrm{F}_{3} \rightarrow$ Force between neutron and 1 proton, nuclear force. $\mathrm{F}_{2} \rightarrow$ Force between protons, there is two acting force, one is nuclear and other is electrostatic. Electrostatic force is repulsive in nature. So, $\mathrm{F}_{2}=$ nuclear force - electrostatic force. Hence, $F_{1}=F_{3}>F_{2}$
