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Vladimir [108]
3 years ago
12

If the work function for a certain metal is 1.8eV, what is the stopping potential for electrons ejected from the metal when ligh

t of wavelength 400nm shines on the metal? And what is the maximum speed of the ejected electrons? ...?
Physics
1 answer:
taurus [48]3 years ago
8 0
First of all the equation you need is this one
<span>E=<span>Ek</span>+ϕ</span> <span><span>Ek</span>=e<span>V0
</span></span>Please remember that the energy of a photon of wavelength lambda is given by <span>E=<span><span>hc</span>λ

</span></span>you can then calculate the maximum kinetic energy since you know the work function of the metal (1.8eV), and the wavelength of the photons (400 nm) (remembering to convert to SI units of joules for the work function). You can then find the maximum velocity of the electrons easy enough.You can obtain the stopping potential usign the next formula
Ek=eV0
<span>Because it is a stopping voltage, V_{0}  and remember it will have the units of volts.
</span>I think with that I can help you a lot. 
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How long does it take a wave to travel 1200 meters with the speed of 3 X 108m/sec?
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Around 3.70 seconds unless traveling through media

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3 years ago
A 5.30 g bullet moving at 963 m/s strikes a 610 g wooden block at rest on a frictionless surface. The bullet emerges, traveling
Tems11 [23]

Answer:

a) V_{wf} = 4.67m/s

b) V = 8.29 m/s

Explanation:

Givens:

The bullet is 5.30g moving at 963m/s and its speed reduced to 426m/s. The wooden block is 610g.

a) From conservation of linear momentum

Pi = Pf

m_{b}V{b_{i} }  + V_{wi}  = m_{w} V_{wf} + m_{b}V_{bf}

where m_{b},V{b_{i} are the mass and the initial velocity of the bullet, m_{w} and V_{wi} are the mass and the initial velocity of the wooden block, and V_{wf} and V_{bf} are the final velocities of the wooden block and the bullet

The wooden block is initial at rest (V_{wi} = 0) this yields

m_{b}V{b_{i} }  = m_{w} V_{wf} + m_{b}V_{bf}

By solving for V_{wf} adn substitute the givens

V_{wf} = \frac{m_{b}V_{bi} - m_{b} V_{bf}    }{m_{W} }

= \frac{5.3(g)(963(m/s)-426(m/s) }{610(g)}

V_{wf} = 4.67m/s

b) The center of mass speed is defined as

V = \frac{m_{b} }{m_{b}+m_{w} } V_{bi}

substituting:

V = \frac{5.3(g)}{5.3(g)+610(g)} X 963(m/s)

V = 8.29 m/s

7 0
3 years ago
Review. Consider the deuterium-tritium fusion reaction with the tritium nucleus at rest:
asambeis [7]

Considering the deuterium-tritium fusion reaction with the tritium nucleus at rest: ¹₂H + ¹₃H → ²₄He + ⁰₁n  the electric potential energy (in electron volts) at this distance is 17.58MeV

<h3>How is the electric potential energy of deuterium-tritium fusion reaction calculated?</h3>

The reaction is  ¹₂H + 1₃H → ²₄He + ⁰₁n

Value of Q = (Mass of ¹₂H + Mass of ¹₃H - Mass of ²₄He- Mass of n) x 931 MeV

Mass of ¹₂H = 2.014102

Mass of ¹₃H = 3.016049

Mass of ²₄He = 4.002603

Mass of n = 1.00867

Therefore Value of Q = [2.014102+3.016049−4.002603−1.00867] × 931 MeV

Therefore Value of Q = 0.01887 × 931 MeV

= 17.58MeV

To learn more about  deuterium-tritium fusion reaction, refer

brainly.com/question/9054784

#SPJ4

7 0
1 year ago
A storage tank containing oil (SG=0.92) is 10.0 meters high and 16.0 meters in diameter. The tank is closed, but the amount of o
blagie [28]

Answer:

a-1 Graph is attached. The relation is linear.

a-2 The corresponding height for 68 kPa Pressure is 7.54 m

a-3 The corresponding weight for 68 kPa Pressure is 1394726kg

b The original height of the column is 5.98 m

Explanation:

Part a

a-1

The graph is attached with the solution. The relation is linear as indicated by the line.

a-2

By the equation

P=\rho \times g \times h

Here

  • P is the pressure which is given as 68 kPa.
  • ρ is the density of the oil whose SG is 0.92. It is calculated as

                                       \rho=S.G \times \rho_{water}\\\rho=0.92 \times 1000 kg/m^3\\\rho=920 kg/m^3\\

  • g is the gravitational constant whose value is 9.8 m/s^2
  • h is the height which is to be calculated

                                        P=\rho \times g \times h\\h=\frac{P}{\rho \times g}\\h=\frac{68 \times 10^3}{920 \times 9.8}\\h=7.54m

So the height of column is 7.54m

a-3

By the relation of volume and density

M=\rho \times V

Here

  • ρ is the density of the oil which is 920 kg/m^3
  • V is the volume of cylinder with diameter 16m calculated as follows

                             V=\pi r^2h\\V=3.14\times (8)^2 \times 7.54\\V=1515.23 m^3

Mass is given as

                             M=\rho \times V\\M=920 \times 1515.23\\M=1394726kg

So the mass of oil leading to 68kPa is 1394726kg

Part b

Pressure variation is given as

                            \Delta P=P_{obs}-P_{atm}\\\Delta P=115-101 kPa\\\Delta P=14 kPa\\

Now corrected pressure is as

P_c=P_g-\Delta P\\P_c=68-14 kPa\\P_c=54 kPa

Finding the value of height for this corrected pressure as

P_c=\rho \times g \times h\\h=\frac{P_c}{\rho \times g}\\h=\frac{54 \times 10^3}{920 \times 9.8}\\h=5.98m

The original height of column is 5.98m

4 0
3 years ago
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