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3 years ago

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6. The table shows the first five energy levels for mercury. A mercury atom makes a transition that emits a photon with a freque

ncy of 4.91 × 1014 Hz. What was the transition? Use GUESS Method!!
Energy levels of mercury:
n energy level eV
1 -10.38
2 -5.74
3 -5.52
4 -4.95
5 -3.71

1 answer:

bezimeni [28]3 years ago

7 0

Now,
E= 2.03 eV
Also,
Ef – Ei = E2 – E5
= -5.74 eV – (–3.71 eV)
= –2.03 eV.
From the conservation of energy, we know that the atom lost energy when that atom emitted the photon. Thus, the transition was from E5 to E2.

The energy levels for an atom are quantized.

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Light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb

masya89 [10]

Answer:

<h2>121ohms</h2>

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

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3 years ago

an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu

Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

$E1= k*q1/(h1-h2)..............(1)$

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1

$[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2$ \\

$E1=[tex]E= 359502+359502\\E=719004 V/m$ V/m[/tex]

similarly electric feild due to negative charge q2=-40

$[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2$ \\

$E2=359502V/m$

Total electric feild E at the aircraft is given as

$E= E1+ E2\\$ ...............(2)

Put values of E1 and E2 in equation2

$\\E=359502+359502\\E= 719004V/m\\$

therefore s Total electric feild E at the aircraft is E= 719004V/m

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2 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

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Silver has a work function of 4.5 eV . Part A What is the longest wavelength of light that will release an electron from a silve

uranmaximum [27]

Answer:

λ = 2.7608 x 10⁻⁷ m = 276.08 nm

Explanation:

The work function of a metallic surface is the minimum amount of photon energy required to release the photo-electrons from the surface of metal. The work function is given by the following formula:

Work Function = hc/λ

where,

Work Function = (4.5 eV)(1.6 x 10⁻¹⁹ J/1 eV) = 7.2 x 10⁻¹⁹ J

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = longest wavelength capable of releasing electron.

Therefore,

7.2 x 10⁻¹⁹ J = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/λ

λ = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(7.2 x 10⁻¹⁹ J)

<u>λ = 2.7608 x 10⁻⁷ m = 276.08 nm</u>

7 0

3 years ago

Is an object moving in uniform circular motion accelerating?

dexar [7]

Yes.

-- 'Acceleration' does NOT mean 'speeding up'.
It means ANY change in the speed OR direction of motion ...
speeding up, slowing down, or turning.

-- If an object is NOT moving in straight line at constant speed,
then its motion is accelerated.

-- In circular motion, or even just going around a curve,
the object is accelerating, because its direction is constantly
changing, even if its speed is constant.

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3 years ago