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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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Answer:

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Explanation:

Given data,

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The formula for time of flight is given by,

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                                u = 15.68 m/s

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3 years ago
An ideal gas Carnot cycle with air in a piston cylinder has a high temperature of 1200 K and a heat rejection at 400 K. During t
lana [24]

Answer:

The specific heat capacity is q_{L}=126.12kJ/kg

The efficiency of the temperature is n_{TH}=0.67

Explanation:

The p-v diagram illustration is in the attachment

T_{H} means high temperature

T_{L} means low temperature

The energy equation :

q_{h} = R* T_{h} in(V_{2}/V_{1})

   =0.287 * 1200 ln(3)

   =0.287*1318.33

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The specific heat capacity:

q_{L}=q_{h}*(T_{L}/T_{H})

q_{L}=378.36 * (400/1200)

q_{L}=378.36 * 0.333

q_{L}=126.12kJ/kg

The efficiency of the temperature will be:

n_{TH}=1 - (T_{L}/T_{H})

n_{TH}=1-(400/1200)

n_{TH}=1-0.333

n_{TH}=0.67

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zaharov [31]

Answer:

okay here is a thing I learned when I was younger in my middle school:

Explanation:

my teacher would tell me that metals are considered a weak metals are on the left side and the good metals are located on the right side because the only way I remembered was the right means it is really strong and the left is weak and not that supportive. but I think that's how I still think it is or other people may have their own opinions. but hope this helped out with your question!

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Answer:

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Explanation:

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