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kirill115 [55]
3 years ago
15

The 488 nm laser is shine on a lithium metal whose work function is 2.9 eV will you be able to see any photoelectrons? if yes wh

at is the kinetic energy of the electrons if no what wavelength of light should be used to see the photoelectrons
Chemistry
1 answer:
nalin [4]3 years ago
3 0

Answer:

There will not be any ejection of photoelectrons

Explanation:

Energy of the photon= hc/λ

Where;

h= Plank's constant

c= speed of light

λ= wavelength of the incident photon

E= 6.6×10^-34 × 3 ×10^8/488 × 10^-9

E= 4.1 ×10^-19 J

Work function of the metal (Wo)= 2.9 eV × 1.6 × 10^-19 = 4.64 × 10^-19 J

There can only be ejected photoelectrons when E>Wo but in this case, E<Wo hence there will not be any ejection of photoelectrons.

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dybincka [34]

Sorry, I won't understand your words.

6 0
3 years ago
Read 2 more answers
Does a large ball of aluminum foil have the same properties as a small ball?
dsp73

Answer:

yes

Explanation:

same properties different size

can i have brainliest pls

5 0
3 years ago
Please help, with step by step work
natali 33 [55]

\qquad ☀️\pink{\bf{ {Answer  = \: \:   85.57g }}}

Molar mass of \bf Cu_2O

\qquad \twoheadrightarrow\sf 63.546 \times 2 +16

\qquad \pink{\twoheadrightarrow\bf 143.092 g}

<u>As we know</u>–

1 mol =\bf 6.02×10^{23} formula units

1 mol\bf Cu_2O = 143.092 g = \bf 6.02×10^{23}formula units

Henceforth –

\bf 3.60×10^{23} formula units \bf Cu_2O–

\qquad \sf :\implies \dfrac{143.092 \times3.60×10^{23  }}{6.02×10^{23}}

\qquad \sf :\implies \dfrac{143.092 \times3.60×\cancel{10^{23  }}}{6.02×\cancel{10^{23}}}

\qquad \pink{:\implies\bf 85.57 g}

5 0
2 years ago
What is NOT a cause of an oil spill?
katrin [286]

Answer:

The answer is #1

Explanation:

oil does not come out of valcanos

6 0
3 years ago
Helpppp pleaseee ill give brainliest
Studentka2010 [4]

Answer:

The answers are in the explanation.

Explanation:

The energy required to convert 10g of ice at -10°C to water vapor at 120°C is obtained per stages as follows:

Increasing temperature of ice from -10°C - 0°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 2.06J/g°C, ΔT is change in temperature = 0°C - -10°C = 10°C and m is mass of ice = 10g

Q = 2.06J/g°C*10°C*10g

Q = 206J

Change from solid to liquid:

The heat of fusion of water is 333.55J/g. That means 1g of ice requires 333.55J to be converted in liquid. 10g requires:

Q = 333.55J/g*10g

Q = 3335.5J

Increasing temperature of liquid water from 0°C - 100°C:

Q = S*ΔT*m

Q is energy, S specific heat of ice = 4.18J/g°C, ΔT is change in temperature = 100°C - 0°C = 100°C and m is mass of water = 10g

Q = 4.18J/g°C*100°C*10g

Q = 4180J

Change from liquid to gas:

The heat of vaporization of water is 2260J/g. That means 1g of liquid water requires 2260J to be converted in gas. 10g requires:

Q = 2260J/g*10g

Q = 22600J

Increasing temperature of gas water from 100°C - 120°C:

Q = S*ΔT*m

Q is energy, S specific heat of gaseous water = 1.87J/g°C, ΔT is change in temperature = 20°C and m is mass of water = 10g

Q = 1.87J/g°C*20°C*10g

Q = 374J

Total Energy:

206J + 3335.5 J + 4180J + 22600J + 374J =

30695.5J =

30.7kJ

5 0
2 years ago
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