Can't copy it so here

Bohr recognized that the electrons in the shells surrounding the nucleus were able to move to different energy levels. if an ele

Over [174]

If an electron absorbs enough energy it would jump to a higher energy level.

4 0

4 years ago

Read 2 more answers

What happens when mix coke and milk

qwelly [4]

ωєℓℓ тнє ρнσѕρнσяι¢ α¢ι∂ мσℓє¢υℓєѕ αттα¢н тσ тнє мσℓє¢υℓєѕ σf тнє мιℓк, αи∂ тнαт ιи¢яєαѕєѕ тнє ∂єиѕιту αи∂ тнєи ѕєρєяαтєѕ тнєм fяσм тнє яєѕт σf тнє ℓιqυι∂ ιи ιт. тнє яємαιи∂єя σf тнє ℓιqυι∂ѕ,иσω нανιиg ℓєѕѕ ∂єиѕιту тнαи тнє ρнσѕρнσяι¢ α¢ι∂ѕ & тнє мιℓк мσℓє¢υℓєѕ, ѕσ ιт ιт иσω fℓσαтѕ σи тσρ.

нσρє ι ¢συℓ∂ нєℓρ уσυ.

7 0

3 years ago

Read 2 more answers

How do I write the formula for: Ferric hydrogen sulphate, Tungsten (V) peroxide, Nitric acid and Aluminum acetate? ...?

hram777 [196]

We write the following substances' chemical formula as follows:

Ferric hydrogen sulphate = Fe(HSO4)3.

Tungsten (V) peroxide = W2(O2)5

Nitric acid = HNO3

Aluminum acetate = C6H9AlO6

Hope this answers the question. Have a nice day.

7 0

3 years ago

A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l

madreJ [45]

Explanation:

From first source, kinetic energy ( $K.E_{1}$ ) ejected is 1 eV and wavelength of light is $\lambda$ .

From second source, kinetic energy ( $K.E_{2}$ ) ejected is 4 eV and wavelength of light is $\frac{\lambda}{2}$ .

Relation between work function, wavelength, and kinetic energy is as follows.

K.E = $\frac{hc}{\lambda} - \phi$

where, h = Plank's constant = $6.63 \times 10^{-34}$ J.s

c = speed of light = $3 \times 10^{8}$ m/s

Also, it is known that 1 eV = $1.6 \times 10^{-19}$ J

Therefore, substituting the values in the above formula as follows.

From first source,

$K.E_{1}$ = $\frac{hc}{\lambda} - \phi$

1 eV = $1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$

$1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (1)

From second source,

$K.E_{2}$ = $\frac{hc}{\lambda} - \phi$

$4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi$

$6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi$ ........... (2)

Now, divide equation (2) by 2. Therefore, it will become

${6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}$

$3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}$ ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

$1.6 \times 10^{-19} = \frac{\phi}{2}$

$\phi$ = $3.2 \times 10^{-19}$

= 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0

4 years ago

2. Explain why atoms of the same element are not all identical.

GarryVolchara [31]

They are not always identical because they have different isotope numbers which develops different neutrons numbers

6 0

3 years ago