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3 years ago

The electron structures of atoms are not involved in the emission of:

1 answer:

7 0

The correct one is gamma rays. Lets go over them one by one.

Gamma rays are rays that arise from gamma decay, a type of radioactive decay. Often, after another decay, the nucleus is still unstable and it gives off energy in the form of gamma rays to stabilize itself. Hence, gamma rays have nothing to do with the electron structure, only with the nucleus of the atom.

X-rays are the product of accelerating electrons, hence only specific atoms can emit a specific energy of X-rays; similarly for the photoelectric phenomenon, the energy which is needed for photoelectrons to be created depends on the electron structure of the atom (in both cases, it is important to see how strong the bond between electron and atom is).
Finally, spectral lines differ depending on the electron structure of the atoms since electrons with different energies absorb different frequencies of light.

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A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while

dem82 [27]

12.8 rad

Explanation:

The angular displacement $\theta$ through which the wheel turned can be determined from the equation below:

$\omega^2 = \omega_0^2 + 2\alpha\theta$ (1)

where

$\omega_0 = 0$

$\omega = 34.7\:\text{rad/s}$

$\alpha = 47.0\:\text{rad/s}^2$

Using these values, we can solve for $\theta$ from Eqn(1) as follows:

$2\alpha\theta = \omega^2 - \omega_0^2$

$\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}$

$\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}$

$\:\:\:\:= 12.8\:\text{rad}$

7 0

2 years ago

The atomic mass of gold is 0.197 kg/mole. how many moles are in 0.566 kg of gold.

iogann1982 [59]

Explanation:

0.566kg *(1mol/0.197 kg)= 2.87 mol gold

note how the units cancel out, if the units do not cancel out (kg/kg=1) then u did something wrong

7 0

3 years ago

Read 2 more answers

Which of the following shows the prefixes in order from largest to smallest? centi, deci, nano

Burka [1]

Deci, Centi, then Nano is the correct order from largest to smallest

5 0

3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0

3 years ago

A 2500-lb vehicle has a drag coefficient of 0.35 and a frontal area of 20 ft2. What is the minimum tractive effort required for

emmasim [6.3K]

Answer:

i put this in the calculator and my answer is 600. hope this helps

Explanation:

4 0

4 years ago