Physics Test - 10

Bohr model is applicable to both the Hydrogen atom and the He⁺ atom.

Hydrogenic atoms are atoms consisting of a nucleus with positive charge Ze⁺ and a single electron, where Z is the proton number. For examples are a hydrogen atom, singly ionized helium, doubly ionized lithium, and so forth.Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two-electron atoms such as helium.

Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

In semiconductors, both electrons and holes are charge carriers and will take part in conduction.

In \(n\)-type semiconductors they are electrons, while in \(p\)-type semiconductors they are holes. The less abundant charge carriers are called minority carriers in \(n\)-type semiconductors they are holes, while in \(p\)-type semiconductors they are electrons.

In a transparent plane sheet, the thickness of the sheet is negligibly small that it allows light to travel through it with negligible deviation.

• The speed of light varies in different media.
• For a lens to act as a transparent sheet when placed in liquid refraction should not take place. So, either its thickness must be negligibly small or the speed of light in the liquid and lens must be the same.
• The former is not possible, as the lens cannot be altered to a thin sheet.
• Thus, the speed of the light in the lens must match that in the liquid. This is possible only if the two media have the same optical density i.e., refractive index.

Thus, the refractive index of the liquid must be equal to that of the lens i.e., 1.49.

Initial flux through the coil,

\(\Phi_{B \text { (initial) }}=B A \cos \theta\)

\(=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 0^{\circ}\)

\(=3 \pi \times 10^{-7} ~Wb\)

Final flux after the rotation,

\(\Phi_{B(\text { final) }}=3.0 \times 10^{-5} \times\left(\pi \times 10^{-2}\right) \times \cos 180^{\circ}\)

\(=-3 \pi \times 10^{-7} ~Wb\)

Therefore, estimated value of the induced emf is,

\(\varepsilon=N \frac{\Delta \Phi}{\Delta t}\)

\(=500 \times\frac{\left(6 \pi \times 10^{-7}\right)}{0.25}\)

\(=3.8 \times 10^{-3} ~V\)

\(I=\frac{\varepsilon}{R}\)

\(=\frac{3.8 \times 10^{-3}}{2}\)

\(=1.9 \times 10^{-3} ~A\)

Given:

Potential difference \(( V )=220 V\)

Power of the bulb \(( P )=40 W\)

We know that:

The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,

\(P=\frac{W}{t}=V I=I^{2} R=\frac{V^{2}}{R}\)

Where, \(V=\) Potential difference, \(R=\) Resistance and \(I=\) current.

The resistance can be calculated as,

\(R=\frac{V^{2}}{P}=\frac{(220)^{2}}{40}=1210 \Omega\)

NAND Implementation:

NAND and NOR gates are known as universal gates. Any one of these gates can be used to implement any kind of logic gate. This kind of feasibility does not exist with other gates i.e. any other gate cannot solely implement all logic gates. For example, AND gate cannot be implemented using an OR gate and vice-versa. The implementation of NAND and NOR gates to generate other logic gates is shown above.

Resistance of coil \(( R )= \frac{V}{I} =\frac{200}{1}=200 \;\Omega\)

With \(AC\) source \(, I=\frac{200}{\sqrt{R}^{2}}\)

Or \(0.5=\frac{200}{\sqrt{R^{2}+X_{L}^{2}}}\)

\(\Rightarrow R^{2}+(2 \pi f L)^{2}=(400)^{2}\)

\(\Rightarrow\left(2 \pi f \times 2 \frac{\sqrt{3}}{\pi}\right)^{2}=(400)^{2}-(200)^{2}\)

\(\Rightarrow\left(2 \pi f \times 2 \frac{\sqrt{3}}{\pi}\right)^2=200 \times 600\)

\(\Rightarrow\left(2 \pi f \times 2 \frac{\sqrt{3}}{\pi}\right)=\sqrt{120000}\)

\(\Rightarrow 4f \sqrt{3}=2 \sqrt{3} \times 100\)

\(\Rightarrow f =\frac{2 \sqrt{3}}{4\sqrt{3}} \times 100\)

\( \Rightarrow f=50~ H z\)

Let \(\mathrm{E_{n}}\) and \(\mathrm{E_{m}}\) be the energies of electron in \(n^{\text {th }}\) and \(m^{\text {th }}\) states.

Then, \(\mathrm{E_{n}-E_{m}=h v_{0}} \ldots(1)\)

In the second case when the atom is moving with a velocity \(\mathrm{v}\). Let \(\mathrm{v^{\prime}}\) be the velocity of atom after emitting the photon. Applying conservation of linear momentum,

\(\mathrm{mv}=\mathrm{mv}^{\prime}+\frac{\mathrm{h\nu}}{\mathrm{c}}\) ( \(\mathrm{m}\) = mass of hydrogen atom)

\(\Rightarrow \mathrm{v}^{\prime}=\left(\mathrm{v}-\frac{\mathrm{h\nu}}{\mathrm{mc}}\right) \ldots(2)\)

Applying conservation of energy

\(\mathrm{E}_{\mathrm{n}}+\frac{1}{2} \mathrm{mv}^{2}=\mathrm{E}_{\mathrm{m}}+\frac{1}{2} \mathrm{mv'}^{2}+\mathrm{h}\nu\)

\(\Rightarrow \mathrm{h} \nu=\left(\mathrm{E}_{\mathrm{n}}-\mathrm{E}_{\mathrm{m}}\right)+\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{v'}^{2}\right)\)

From equation (1) and (2)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\left(\mathrm{v}-\frac{\mathrm{h} \nu}{\mathrm{mc}}\right)^{2}\right]\)

\(=\mathrm{h}\nu_{0}+\frac{1}{2} \mathrm{~m}\left[\mathrm{v}^{2}-\mathrm{v}^{2}-\frac{\mathrm{h}^{2} \nu ^{2}}{\mathrm{~m}^{2} \mathrm{c}^{2}}+\frac{2 \mathrm{h\nu v}}{\mathrm{mc}}\right]\)

\(=\mathrm{h} \nu_{0}+\frac{\mathrm{h} \nu \mathrm{v}}{\mathrm{c}}-\frac{\mathrm{h}^{2} \nu^{2}}{2 \mathrm{mc}^{2}}\)

Here the term is \(\frac{\mathrm{h}^{2}\nu^{2}}{2 \mathrm{mc}^{2}}\) is very small. So, can be neglected.

\(\therefore \mathrm{h \nu=h \nu_{0}+\frac{h \nu v}{c}}\)

\(\Rightarrow \nu=\nu_0 +\frac{\nu\mathrm{v}}{c}\)

\(\Rightarrow \nu_{0}=\nu\left(1-\frac{\mathrm{v}}{\mathrm{c}}\right)\)

Limitations of the Bohr model:

1. The Bohr model is applicable to hydrogenic atoms. It cannot be extended even to mere two-electron atoms such as helium.

• The analysis of atoms with more than one electron was attempted on the lines of Bohr’s model for hydrogenic atoms but did not meet with any success. 
• The formulation of the Bohr model involves electrical force between the positively charged nucleus and electron. It does not include the electrical forces between electrons which necessarily appear in multi-electron atoms.

2. Bohr’s model correctly predicts the frequencies of the light emitted by hydrogenic atoms, the model is unable to explain the relative intensities of the frequencies in the spectrum.

Given:

\(P _{1}=200 W\)

\(P _{2}=100 W\)

\(V =100 V\)

We know that:

The rate at which the electric energy is dissipated or consumed is termed as electric power.

The electric power is given as,

\(P=I V=I^{2} R=\frac{V^{2}}{R}\)

Where, \(P =\) electric power, \(V =\) voltage, \(I =\) current and \(R =\) resistance

\(R=\frac{V^{2}}{P}\)

\(R \propto \frac{1}{P} \quad \ldots(1)\)

For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.

So, the bulb which has more power will have low resistance, therefore the \(100 W\) bulb will have more resistance compared to the \(200 W\) bulb.

The heat dissipated by the bulb is given as,

\(H = I ^{2} R\)

Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.

Since the resistance of the \(100 W\) bulb is more, so the heat dissipation of the \(100 W\) bulb will be more.

The brightness of the bulb depends on the heat dissipation by the bulb, so the \(100 W\) bulb will have more brightness.