Which element gains two electrons to fill its outermost energy level?

Two atoms form a stable chemical bond when their potential energies are ___ than their potential energies as separate particles

musickatia [10]

It is a reactant hope it helped

3 0

3 years ago

If you hit the surface of Iron with a photon of energy and find that the ejected electron has a wavelength of .75 nm, what is th

lubasha [3.4K]

Answer:

The wavelength of the incoming photon is 172.8 nm

Explanation:

The wavelength of the incoming photon can be calculated with the photoelectric equation:

$KE = h\frac{c}{\lambda_{p}} - \phi$ (1)

Where:

KE: is the kinetic energy of the electron

h: is Planck's constant = 6.62x10⁻³⁴ J.s

c: is the speed of light = 3.00x10⁸ m/s

$\lambda_{p}$ : is the wavelength of the photon =?

Φ: is the work function of the surface (Iron) = 4.5 eV

The kinetic energy of the electron is given by:

$KE = \frac{p^{2}}{2m} = \frac{(\frac{h}{\lambda_{e}})^{2}}{2m}$ (2)

Where:

p: is the linear momentum = h/λ

m: is the electron's mass = 9.1x10⁻³¹ kg

$\lambda_{e}$ : is the wavelength of the electron = 0.75 nm = 0.75x10⁻⁹ m

Hence, the wavelength of the photon is:

$\frac{(\frac{h}{\lambda_{e}})^{2}}{2m} = h\frac{c}{\lambda_{p}} - \phi$

$\lambda_{p} = \frac{hc}{\frac{h^{2}}{2m\lambda_{e}^{2}} + \phi} = \frac{6.62 \cdot 10^{-34} J.s*3.00\cdot 10^{8} m/s}{\frac{(6.62 \cdot 10^{-34} J.s)^{2}}{2*9.1 \cdot 10^{-31} kg*(0.75 \cdot 10^{-9} m)^{2}} + 4.5 eV*\frac{1.602 \cdot 10^{-19} J}{1 eV}} = 1.728 \cdot 10^{-7} m = 172.8 nm$

Therefore, the wavelength of the incoming photon is 172.8 nm.

I hope it helps you!

3 0

3 years ago

Answer true or false and please explain why.

AlekseyPX

true because of this When water vapor in the atmosphere loses heat and cools down, condensation happens. As the water vapor cools down and condenses, it attaches to small particles of dust floating in the atmosphere, forming tiny liquid water droplets.

8 0

3 years ago

I NEED HELP!!

shusha [124]

Answer:

1 B

2A

#D

3B

1D

2B

3C

4A

A,C,A,B

C,A,C,A

C,A,A,B

Explanation:

Easy ive done this brings back memories

7 0

3 years ago

How many atoms of K are present in 195.49 grams of K? 3.0110 × 1024 6.0220 × 1024 1.1772 × 1026 4.5797 × 1027

Ierofanga [76]

<h2>Hello!</h2>

The answer is: $3.0110x10^{24}$

First, we need to look for the K molecular mass:

$K=39,098 g/mol$

Also, we need to remember that 1 mol of a compound or element is equal to $6,022x10^{23}$ of that compound.

From the statement, we are asked to calculate how many atoms of K are present in 195,49 of K.

So, doing the calculations, we have:

$195,49g*\frac{1mol}{39,08}*\frac{6,022x10^{23} atoms}{1 mol K} =3,0110x10^{24} \\$

So, we have $3.0110x10^{24}$ atoms of K in 195,49 grams of K.

Have a nice day!

3 0

3 years ago