Which type of element seeks to gain electrons? Question 23 options: A. Non-Metals B. Metalloids C. Isotopes D. Metals E. Noble G

The amount of energy necessary to remove an electron from an atom is a quantity called the ionization energy, Ei. This energy ca

Keith_Richards [3271]

Answer:

The resulting value is $E_i = 1.5596 *10^{-18} \ J$ .

Explanation:

The question specifies that

The wavelength is $\lambda = 48.2 nm = 48.2 *10^{- 9 }\ m$ .

The velocity is $v = 2.371*10^6 \ m/s$ .

The mass of the electron is $m_e = 9.109*10^{-31} \ kg$ .

The energy of the incoming light is typically depicted mathematically as

$E = \frac{h * c}{\lambda}$ .

Here, c represents the speed of light with the value $c = 3.0 *10^{8} \ m/s$ .

h stands for Planck's constant with a value of $h = 6.62607015 * 10^{-34 } J\cdot s$ .

Thus,

$E = \frac{6.62607015 * 10^{-34 }* 3.0 *10^{8}}{48.2 *10^{- 9 }}$

=> $E = 4.12 *10^{-18} \ J$ .

Typically, kinetic energy is represented as

$E_k = \frac{1}{2} * m_e * v^2$

=> $E_k = \frac{1}{2} * 9.109*10^{-31} * (2.371*10^6 )^2$ .

=> $E_k = 2.56 *0^{-18} \ J$ .

The ionization energy is generally expressed mathematically as

$E_i = 4.12 *10^{-18} - 2.56 *0^{-18}$

=> $E_i = 1.5596 *10^{-18} \ J$ .

8 0

2 months ago

Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio

Yuliya22 [3333]

Answer:

B ) Ascend using my buddy's alternative air source / perform an emergency swimming ascent

Explanation:

From the details given, it's clear the buddy is nearby and thus the alternative air source can be utilized easily. Furthermore, the diver is closer to the water's surface than their buddy, making a controlled emergency ascent preferred in this situation.

5 0

3 months ago

Read 2 more answers

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

kicyunya [3294]

Final temperature to determine: Given the following details, the calculations proceed as follows: Mass of the silver ring is m = 4 g, initial temperature is presented, and the heat released is Q = -18 J (indicating heat loss). The specific heat of silver is considered next to find the final temperature.

5 0

2 months ago

An electron moves in a region where the magnetic field is uniform and has a magnitude of 80 μT. The electron follows a helical p

Softa [3030]

Answer:

3.4 x 10⁴ m/s

Explanation:

Analyze the circular path of the electron

B = magnetic field = 80 x 10⁻⁶ T

m = mass of an electron = 9.1 x 10⁻³¹ kg

v = speed in the radial direction

r = radius of the circular trajectory = 2 mm = 0.002 m

q = charge of an electron = 1.6 x 10⁻¹⁹ C

For the electron’s circular movement

qBr = mv

(1.6 x 10⁻¹⁹) (80 x 10⁻⁶) (0.002) = (9.1 x 10⁻³¹) v

v = 2.8 x 10⁴ m/s

Now, consider the electron's movement in a straight line:

v' = speed in linear motion

x = distance traveled horizontally = 9 mm = 0.009 m

t = duration = $\frac{2\pi m}{qB}$ = $\frac{2\pi (9.1\times 10^{-31})}{(1.6\times 10^{^{-19}})(80\times 10^{-6})}$ = 4.5 x 10⁻⁷ sec

Using the formula

x = v' t

0.009 = v' (4.5 x 10⁻⁷)

v' = 20000 m/s

v' = 2 x 10⁴ m/s

The resultant speed is given by

V = sqrt(v² + v'²)

V = sqrt((2.8 x 10⁴)² + (2 x 10⁴)²)

v = 3.4 x 10⁴ m/s

6 0

2 months ago

A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i

Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

$a=\frac{v^2}{R}$

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = $9.81\ m/s^2$

v = ?

R = 13.2 m

Por lo tanto,

$9.81=\frac{v^2}{13.2}$

$v^2=9.81\times {13.2}$

v = 11.4 m/s

8 0

3 months ago