NEET Test Series from KOTA - 10 Papers In MS WORD
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Structure of Atom
238812
The exact path of electron in any orbital cannot be determined. The above statement is based on #[Qdiff: Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
1 Hund's Rule
2 Bohr's Rule
3 Uncertainty Principal
4 Aufbau Principal
Explanation:
According to Heisenberg uncertainty principle, the position and the velocity of an object cannot both be measured exactly at the same time. Hund's Rule- Every orbitals in a subshell is singly occupied with one electron before any one orbital is doubly occupied, Aufbau Principle- Electrons fill lower energy atomic orbitals before filling higher energy ones.
Structure of Atom
238824
A golf-ball weigh $40.0 \mathrm{~g}$. If it is moving with a velocity of $20.0 \mathrm{~ms}^{-1}$, it's de-Broglie wave length is
238826
Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength $550 \mathrm{~nm}$ is $\longrightarrow$, if $100 \mathrm{~V}$ and $1 \mathrm{~A}$ is supplied for one hour.
1 $1 \times 10^{24}$
2 $5 \times 10^{24}$
3 $1 \times 10^{23}$
4 $5 \times 10^{23}$
5 $5 \times 10^{22}$
Explanation:
Given that, $\begin{gathered} \lambda=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m} \\ \mathrm{i}=1 \mathrm{~A}, \mathrm{~V}=100 \mathrm{~V} \\ \therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{550 \times 10^{-9}}=0.036 \times 10^{-17} \end{gathered}$ So, each proton carries energy, $\mathrm{E}=0.036 \times 10^{-17}$ No. of photons $=\frac{\text { total energy output in one second }}{\text { energy carried per photon }}$ $\begin{aligned} & \mathrm{n}=\frac{100 \times 3600}{0.036 \times 10^{-17}} \\ & \mathrm{n}=1 \times 10^{24} \end{aligned}$
Kerala-CEE-2019
Structure of Atom
238827
The ratio of de-Broglie wavelengths for electrons accelerated through $200 \mathrm{~V}$ and $50 \mathrm{~V}$ is :
1 $1: 2$
2 $2: 1$
3 $3: 10$
4 $10: 3$
Explanation:
As we know, $\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2$ So, $\quad v=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Ans, $\quad \mathrm{mv}=\sqrt{2 \mathrm{meV}}$ According to de Broglie equation- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ or $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ When it is accelerated through 200 volt $\lambda_1=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{200}}$ When it is accelerated through 50 volt $\lambda_2=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{50}}$ On solving equation (i) and (ii), we get $\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{50}}{\sqrt{200}}=\frac{1}{2}$
238812
The exact path of electron in any orbital cannot be determined. The above statement is based on #[Qdiff: Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
1 Hund's Rule
2 Bohr's Rule
3 Uncertainty Principal
4 Aufbau Principal
Explanation:
According to Heisenberg uncertainty principle, the position and the velocity of an object cannot both be measured exactly at the same time. Hund's Rule- Every orbitals in a subshell is singly occupied with one electron before any one orbital is doubly occupied, Aufbau Principle- Electrons fill lower energy atomic orbitals before filling higher energy ones.
Structure of Atom
238824
A golf-ball weigh $40.0 \mathrm{~g}$. If it is moving with a velocity of $20.0 \mathrm{~ms}^{-1}$, it's de-Broglie wave length is
238826
Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength $550 \mathrm{~nm}$ is $\longrightarrow$, if $100 \mathrm{~V}$ and $1 \mathrm{~A}$ is supplied for one hour.
1 $1 \times 10^{24}$
2 $5 \times 10^{24}$
3 $1 \times 10^{23}$
4 $5 \times 10^{23}$
5 $5 \times 10^{22}$
Explanation:
Given that, $\begin{gathered} \lambda=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m} \\ \mathrm{i}=1 \mathrm{~A}, \mathrm{~V}=100 \mathrm{~V} \\ \therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{550 \times 10^{-9}}=0.036 \times 10^{-17} \end{gathered}$ So, each proton carries energy, $\mathrm{E}=0.036 \times 10^{-17}$ No. of photons $=\frac{\text { total energy output in one second }}{\text { energy carried per photon }}$ $\begin{aligned} & \mathrm{n}=\frac{100 \times 3600}{0.036 \times 10^{-17}} \\ & \mathrm{n}=1 \times 10^{24} \end{aligned}$
Kerala-CEE-2019
Structure of Atom
238827
The ratio of de-Broglie wavelengths for electrons accelerated through $200 \mathrm{~V}$ and $50 \mathrm{~V}$ is :
1 $1: 2$
2 $2: 1$
3 $3: 10$
4 $10: 3$
Explanation:
As we know, $\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2$ So, $\quad v=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Ans, $\quad \mathrm{mv}=\sqrt{2 \mathrm{meV}}$ According to de Broglie equation- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ or $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ When it is accelerated through 200 volt $\lambda_1=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{200}}$ When it is accelerated through 50 volt $\lambda_2=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{50}}$ On solving equation (i) and (ii), we get $\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{50}}{\sqrt{200}}=\frac{1}{2}$
238812
The exact path of electron in any orbital cannot be determined. The above statement is based on #[Qdiff: Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
1 Hund's Rule
2 Bohr's Rule
3 Uncertainty Principal
4 Aufbau Principal
Explanation:
According to Heisenberg uncertainty principle, the position and the velocity of an object cannot both be measured exactly at the same time. Hund's Rule- Every orbitals in a subshell is singly occupied with one electron before any one orbital is doubly occupied, Aufbau Principle- Electrons fill lower energy atomic orbitals before filling higher energy ones.
Structure of Atom
238824
A golf-ball weigh $40.0 \mathrm{~g}$. If it is moving with a velocity of $20.0 \mathrm{~ms}^{-1}$, it's de-Broglie wave length is
238826
Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength $550 \mathrm{~nm}$ is $\longrightarrow$, if $100 \mathrm{~V}$ and $1 \mathrm{~A}$ is supplied for one hour.
1 $1 \times 10^{24}$
2 $5 \times 10^{24}$
3 $1 \times 10^{23}$
4 $5 \times 10^{23}$
5 $5 \times 10^{22}$
Explanation:
Given that, $\begin{gathered} \lambda=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m} \\ \mathrm{i}=1 \mathrm{~A}, \mathrm{~V}=100 \mathrm{~V} \\ \therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{550 \times 10^{-9}}=0.036 \times 10^{-17} \end{gathered}$ So, each proton carries energy, $\mathrm{E}=0.036 \times 10^{-17}$ No. of photons $=\frac{\text { total energy output in one second }}{\text { energy carried per photon }}$ $\begin{aligned} & \mathrm{n}=\frac{100 \times 3600}{0.036 \times 10^{-17}} \\ & \mathrm{n}=1 \times 10^{24} \end{aligned}$
Kerala-CEE-2019
Structure of Atom
238827
The ratio of de-Broglie wavelengths for electrons accelerated through $200 \mathrm{~V}$ and $50 \mathrm{~V}$ is :
1 $1: 2$
2 $2: 1$
3 $3: 10$
4 $10: 3$
Explanation:
As we know, $\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2$ So, $\quad v=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Ans, $\quad \mathrm{mv}=\sqrt{2 \mathrm{meV}}$ According to de Broglie equation- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ or $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ When it is accelerated through 200 volt $\lambda_1=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{200}}$ When it is accelerated through 50 volt $\lambda_2=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{50}}$ On solving equation (i) and (ii), we get $\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{50}}{\sqrt{200}}=\frac{1}{2}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Structure of Atom
238812
The exact path of electron in any orbital cannot be determined. The above statement is based on #[Qdiff: Easy, QCat: Theory Based, examname: Shift-I , AP- EAPCET-07-09-2021]#
1 Hund's Rule
2 Bohr's Rule
3 Uncertainty Principal
4 Aufbau Principal
Explanation:
According to Heisenberg uncertainty principle, the position and the velocity of an object cannot both be measured exactly at the same time. Hund's Rule- Every orbitals in a subshell is singly occupied with one electron before any one orbital is doubly occupied, Aufbau Principle- Electrons fill lower energy atomic orbitals before filling higher energy ones.
Structure of Atom
238824
A golf-ball weigh $40.0 \mathrm{~g}$. If it is moving with a velocity of $20.0 \mathrm{~ms}^{-1}$, it's de-Broglie wave length is
238826
Maximum number of photons emitted by a bulb capable of producing monochromatic light of wavelength $550 \mathrm{~nm}$ is $\longrightarrow$, if $100 \mathrm{~V}$ and $1 \mathrm{~A}$ is supplied for one hour.
1 $1 \times 10^{24}$
2 $5 \times 10^{24}$
3 $1 \times 10^{23}$
4 $5 \times 10^{23}$
5 $5 \times 10^{22}$
Explanation:
Given that, $\begin{gathered} \lambda=550 \mathrm{~nm}=550 \times 10^{-9} \mathrm{~m} \\ \mathrm{i}=1 \mathrm{~A}, \mathrm{~V}=100 \mathrm{~V} \\ \therefore \mathrm{E}=\frac{\mathrm{hc}}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{550 \times 10^{-9}}=0.036 \times 10^{-17} \end{gathered}$ So, each proton carries energy, $\mathrm{E}=0.036 \times 10^{-17}$ No. of photons $=\frac{\text { total energy output in one second }}{\text { energy carried per photon }}$ $\begin{aligned} & \mathrm{n}=\frac{100 \times 3600}{0.036 \times 10^{-17}} \\ & \mathrm{n}=1 \times 10^{24} \end{aligned}$
Kerala-CEE-2019
Structure of Atom
238827
The ratio of de-Broglie wavelengths for electrons accelerated through $200 \mathrm{~V}$ and $50 \mathrm{~V}$ is :
1 $1: 2$
2 $2: 1$
3 $3: 10$
4 $10: 3$
Explanation:
As we know, $\mathrm{eV}=\frac{1}{2} \mathrm{mv}^2$ So, $\quad v=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$ Ans, $\quad \mathrm{mv}=\sqrt{2 \mathrm{meV}}$ According to de Broglie equation- $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ or $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ When it is accelerated through 200 volt $\lambda_1=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{200}}$ When it is accelerated through 50 volt $\lambda_2=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}} \cdot \sqrt{50}}$ On solving equation (i) and (ii), we get $\frac{\lambda_1}{\lambda_2}=\frac{\sqrt{50}}{\sqrt{200}}=\frac{1}{2}$