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3 years ago

How does the quantum mechanical model of the atom describe electrons

1 answer:

5 0

The quantum mechanical model describes the allowed energies an electron can have. It also describes how likely it is to find the electrons in various locations around an atom's nucleus.

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One electron collides elastically with a second electron initially at rest. After the collision, the radii of their trajectories

ch4aika [34]

Answer:

114.92749 keV

Explanation:

r = Radius of trajectory

m = Mass of electron = $9.11\times 10^{-31}\ kg$

B = Magnetic field = 0.044 T

q = Charge of electron = $1.6\times 10^{-19}\ C$

The centripetal force and the magnetic forces are conserved

$m\frac{v^2}{r}=Bqv\\\Rightarrow v=\frac{Bqr}{m}$

Velocity of first electron

$v=\frac{Bqr_1}{m}\\\Rightarrow v=\frac{0.044\times 1.6\times 10^{-19}\times 0.01}{9.11\times 10^{-31}}\\\Rightarrow v_1=77277716.79473\ m/s$

Velocity of second electron

$v=\frac{Bqr_2}{m}\\\Rightarrow v_2=\frac{0.044\times 1.6\times 10^{-19}\times 0.024}{9.11\times 10^{-31}}\\\Rightarrow v_2=185466520.30735\ m/s$

Total kinetic energy is given by

$K=K_1+K_2\\\Rightarrow K=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2\\\Rightarrow K=\frac{1}{2}m(v_1^2+v_2^2)\\\Rightarrow K=\frac{1}{2}\times 9.11\times 10^{-31}(77277716.79473^2+185466520.30735^2)\\\Rightarrow K=1.83884\times 10^{-14}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$1.83884\times 10^{-14}\ J=1.83884\times 10^{-14}\times \frac{1}{1.6\times 10^{-19}}\ eV\\ =114927.49\ ev=114.92749\ keV$

The energy of incident electron is 114.92749 keV

5 0

3 years ago

WHATS THE CORRECT ANSWER ANSWER ASAP

Harman [31]

Answer:

speed and velocity

Explanation:

8 0

1 year ago

Read 2 more answers

How does the motion of the molecules change when you increase the temperature

borishaifa [10]

Answer:

They speed up as temperature increases.

Explanation:

As temperatures increases within a molecule the particles will speed up.

6 0

2 years ago

A laboratory technician drops a 72.0 g sample of unknown solid material, at a temperature of 80.0°C, into a calorimeter. The cal

Natalija [7]

Answer : The specific heat of unknown sample is, $8748.78J/kg^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-[q_2+q_3]$

$m_1\times c_1\times (T_f-T_1)=-[m_2\times c_2\times (T_f-T_2)+m_3\times c_3\times (T_f-T_2)]$

where,

$c_1$ = specific heat of unknown sample = ?

$c_2$ = specific heat of water = $4186J/kg^oC$

$c_3$ = specific heat of copper = $390J/kg^oC$

$m_1$ = mass of unknown sample = 72.0 g = 0.072 kg

$m_2$ = mass of water = 203 g = 0.203 kg

$m_2$ = mass of copper = 187 g = 0.187 kg

$T_f$ = final temperature of calorimeter = $39.4^oC$

$T_1$ = initial temperature of unknown sample = $80.0^oC$

$T_2$ = initial temperature of water and copper = $11.0^oC$

Now put all the given values in the above formula, we get

$0.072kg\times c_1\times (39.4-80.0)^oC=-[(0.203kg\times 4186J/kg^oC\times (39.4-11.0)^oC)+(0.187kg\times 390J/kg^oC\times (39.4-11.0)^oC)]$

$c_1=8748.78J/kg^oC$

Therefore, the specific heat of unknown sample is, $8748.78J/kg^oC$

7 0

3 years ago

What happens when heat is transferred by convection

Dimas [21]

Convection is heat transfer by mass motion of a fluid such as air or water when the heated fluid is caused to move away from the source of heat, carrying energy with it.

3 0

3 years ago