Ask question

Ask question

All categories

3 years ago

7

Imagine that an electron in an excited state in a nitrogen molecule decays to its ground state, emitting a photon with a frequen

cy of 8.88×1014 Hz . What is the change in energy, ΔE, that the electron undergoes to decay to its ground state?

1 answer:

wariber [46]3 years ago

5 0

Answer: 5.884×10^-19 J

Explanation:

The change in energy will be the product of the planck constant and the frequency of radiation.

ΔE = hf

Where h = planck constant = 6.626×10^-34 m²kg/s

f = 8.88×10^14 Hz

ΔE = 6.626×10^-34 × 8.88×10^14 = 58.84×10^-20 = 5.884×10^-19 J

You might be interested in

A 1.0 kg object is attached to a 0.50 m string. It is twirled in a horizontal circle above the ground at a speed of 5.0 m/s.

7nadin3 [17]

Given that,

Mass of the object, m = 1 kg

It moves in a circle of radius 0.5 m with a speed of 5 m/s

To find,

The direction of the acceleration.

Solution,

Whenever an object moves in a circular path, the only force that acts on its is centripetal force which is given by the formula as follows :

$F=\dfrac{mv^2}{r}$

The centripetal acceleration acts in the direction of force. It acts along the radius of the circular path. In this figure, the direction of the acceleration is shown at point d.

4 0

3 years ago

As wavelength decreases, frequency and energy _________________;

earnstyle [38]

Answer:

Increase

Explanation:

The best way for me to visualize the relation between wavelength, frequency, and energy is to think about actual ocean waves. Wavelength is a measure of the distance between two equivalent points on consecutive waves (think wave peak to wave peak). Lets say you are building a sand castle and want to see how many waves hit your castle over a period of 10 seconds. If the distance between each wave is 10 ft and the wave is traveling at 1 foot per second then you will only have one wave hit your castle. If the wavelength is 1/2 that (5 ft) then you will have 2 waves hit your castle in the same amount of time. This is the same concept behind waves in physics. The smaller the distance between each wave, the more waves and therefore more energy that will be delivered.

4 0

3 years ago

A jet plane traveling 1890 km/h (525 m/s) pulls out of a dive by moving in an arc of radius 5.20 km. What is the plane's acceler

Tema [17]

Answer:

Acceleration of the plane, a = 5.4 g

Explanation:

It is given that,

Speed of the jet plane, v = 1890 km/h = 525 m/s

Radius of the arc, r = 5.20 km = 5200 m

The plane is moving in the circular path, the centripetal acceleration will act on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(525\ m/s)^2}{5200\ m}$

$a=53.004\ m/s^2$

We know that, the value of g is, $g=9.8\ m/s^2$

$\dfrac{a}{g}=\dfrac{53.004\ m/s^2}{9.8\ m/s^2}=5.4$

$a=5.4\times g$

So, the acceleration of the plane is 5.4 g. Hence, this is the required solution.

6 0

3 years ago

Read 2 more answers

The Moon requires about 1 month (0.08 year) to orbit Earth. Its distance from us is about 400,000 km (0.0027 AU). Use Kepler’s t

dem82 [27]

Answer:

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

Explanation:

As per Kepler's III law we know that time period of revolution of satellite or planet is given by the formula

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

now for the time period of moon around the earth we can say

$T_1 = 2\pi\sqrt{\frac{r_1^3}{GM_e}}$

here we know that

$T_1 = 0.08 year$

$r_1 = 0.0027 AU$

$M_e$ = mass of earth

Now if the same formula is used for revolution of Earth around the sun

$T_2 = 2\pi\sqrt{\frac{r_2^3}{GM_s}}$

here we know that

$r_2 = 1 AU$

$T_2 = 1 year$

$M_s$ = mass of Sun

now we have

$\frac{T_2}{T_1} = \sqrt{\frac{r_2^3 M_e}{r_1^3 M_s}}$

$\frac{1}{0.08} = \sqrt{\frac{1 M_e}{(0.0027)^3M_s}}$

$12.5 = \sqrt{(5.08 \times 10^7)\frac{M_e}{M_s}}$

$\frac{M_e}{M_s} = 3.07 \times 10^{-6}$

4 0

3 years ago

Do I fill this chart right? Can you please check if I am wrong or not?

luda_lava [24]

Physical properties of water:

Color: No color/colorless.

Odor: Odorless.

Taste: Tasteless.

Density: 1 g/ml.

Boiling point: 100°C.

Freezing/Melting point: 0°C.

State or phase:

Vapor: Gas.

Room temperature: Liquid.

Ice: Solid.

Hence, water is a colorless, odorless, and tasteless substance whose density is 1 g/ml. It boils at 100°C and melts at 0°C. In the vapor phase, it is in a gaseous state, at room temperature, it exists in a liquid state, and when its freezes (ice) it's in the solid state.

7 0

2 years ago