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Leona [35]
2 years ago
14

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

he work function W0 of this metal? Express your answer in electron volts.
Physics
1 answer:
Veseljchak [2.6K]2 years ago
5 0

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

W_{o}=\frac{hc}{\lambda }

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }

W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

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Flapping flight is very energy intensive. A wind tunnel test
steposvetlana [31]

Answer:

The metabolic power for starting flight=134.8W/kg

Explanation:

We are given that

Mass of starling, m=89 g=89/1000=0.089 kg

1 kg=1000 g

Power, P=12 W

Speed, v=11 m/s

We have to find the metabolic power for starting flight.

We know that

Metabolic power for starting flight=\frac{P}{m}

Using the formula

Metabolic power for starting flight=\frac{12}{0.089}

Metabolic power for starting flight=134.8W/kg

Hence, the metabolic power for starting flight=134.8W/kg

4 0
3 years ago
Define energy and provide examples of potential and kinetic energy
In-s [12.5K]

Energy is the capacity for doing work..

Kinetic energy - Moving car

Potential energy - flowing water up the hill

6 0
3 years ago
What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?
fiasKO [112]
Momentum = mv
where m is the mass of an electron and v is the velocity of the electron.

v = momentum ÷ m
   = (1.05×10∧-24)÷(9.1×10∧-31) = 1,153,846.154 m/s

kinetic energy = (mv∧2)÷2
                       = (9.1×10∧-31 × 1,153,846.154∧2) ÷2
                      = (1.21154×10∧-18) ÷ 2
                      = 6.05769×10∧-19 J
4 0
3 years ago
Read 2 more answers
A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear
Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

f_o = f_s\frac {(v+vs)}{v} and f_o = f_s\frac {(v-vs)}{v} when moving towards observer

With f_o of 76 then taking speed in air as 343 m/s we have

76 = f_s\times\frac {(343-vs)}{343}

f_s=\frac {343\times 76}{343-v_s}

Similarly, with f_o of 65 we have

65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}

Now

f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}

v_s=27.76 m/s

Substituting the above into  any of the first two equations then we obtain

f_s=70.07 Hz

4 0
3 years ago
Help me please I need help?
sweet [91]
Your answer is c. homologous structures
4 0
3 years ago
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