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3 years ago

8

A quantum-mechanical system initially in its ground level absorbs a photon and ends up in the first excited state. The system th

en absorbs a second photon and ends up in the second excited state. For which of the following systems does the second photon have a longer wavelength than the first one?
(a) a harmonic oscillator;

(b) a hydrogen atom;

(c) a particle in a box.

1 answer:

Ludmilka [50]3 years ago

5 0

Answer:

(b) a hydrogen atom

Explanation:

Wavelength is inversely proportional to energy. So we have to find the system for which the second photon has smaller energy than the first one.

For the harmonic oscillator, the energy level spacing remains the same as the quantum number increases.

For the hydrogen atom, the energy level spacing decreases as the quantum number increases.

For the particle in a box, the energy level spacing increases as the quantum number increases.

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3 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

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Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

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4 years ago

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3 years ago

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vladimir2022 [97]

To solve this problem it is necessary to apply the concepts related to the adiabatic process that relate the temperature and pressure variables

Mathematically this can be determined as

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

Where

$T_1 =$ Temperature at inlet of turbine

$T_2 =$ Temperature at exit of turbine

$P_1 =$ Pressure at exit of turbine

$P_2 =$ Pressure at exit of turbine

The steady flow Energy equation for an open system is given as follows:

$m_i = m_0 = m$

$m(h_i+\frac{V_i^2}{2}+gZ_i)+Q = m(h_0+\frac{V_0^2}{2}+gZ_0)+W$

Where,

m = mass

$m_i$ = mass at inlet

$m_0$ = Mass at outlet

$h_i$ = Enthalpy at inlet

$h_0$ = Enthalpy at outlet

W = Work done

Q = Heat transferred

$V_i$ = Velocity at inlet

$V_0$ = Velocity at outlet

$Z_i$ = Height at inlet

$Z_0$ = Height at outlet

For the insulated system with neglecting kinetic and potential energy effects

$h_i = h_0 + W$

$W = h_i -h_0$

Using the relation T-P we can find the final temperature:

$\frac{T_2}{T_1} = (\frac{P_2}{P_1})^{(\frac{\gamma-1}{\gamma})}$

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So:

$W = h_i -h_0$

$W = C_p (T_1-T_2)$

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4 years ago