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2 years ago

6

A photon with wavelength 11. 0 nm is absorbed when an electron in a three-dimensional cubical box makes a transition from the gr

ound state to the second excited state

1 answer:

const2013 [10]2 years ago

3 0

For a photon with wavelength 11. 0 nm, the side length LL of the box is mathematically given as

L=1.414*10^{-27}m

<h3>What is the side length LL of the box?</h3>

Generally, the equation for the change in energy is mathematically given as

$dE=\frac{hc}{\lambda}$

Therefore

$\frac{hc}{\lambda}=\frac{6h^2}{8ml}\\\\L=\sqrt{6h\lambda}{8mc}\\\\L=\sqrt{\frac{6(6.626*10^{-34})*11.0*10^{-9}}{8(9.11*10^31)(3*10^8)}}$

L=1.414*10^{-27}m

In conclusion, the side length LL of the box is

L=1.414*10^{-27}m

Read more about Mesurement

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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

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3 years ago

A bullet is fired horizontally from a handgun at a target 100.0 m away. If the initial speed of the bullet as it leaves the gun

Lera25 [3.4K]

Answer:

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Explanation:

Given that,

Target distance = 100.0 m

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We need to calculate the total time

Using formula of time

$t=\dfrac{d}{v}$

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$t=\dfrac{100.0}{300}$

$t=0.33\ sec$

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Initial velocity of bullet in vertical direction = 0 m/s

We need to calculate the vertically distance

Using equation of motion

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Answer:

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Explanation:

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