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dexar [7]
3 years ago
5

Light that has a wavelength of 600 nm strikes a metal surface, and a stream of electrons is ejected from the surface. If light o

f wavelength 500 nm strikes the surface, the maximum kinetic energy of the electrons emitted from the surface will be
Smaller

The Same

Greater

Smaller by a factor of 5/6

Unmeasurable
Physics
1 answer:
Anni [7]3 years ago
8 0

Answer:

The maximum kinetic energy of the electrons emitted from the surface will be greater.

Option 3 is correct.

Explanation:

Given that,

Wavelength \lambda_{1}=600\ nm

Wavelength \lambda_{2}=500\ nm

We need to calculate the energy for first wavelength

Using formula of energy

E=hf

E=h\dfrac{c}{\lambda}

Put the value into the formula

E_{1}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{600\times10^{-9}}

E_{1}=3.33\times10^{-19}\ J

We need to calculate the energy for second wavelength

Put the value into the formula

E_{2}=6.67\times10^{-34}\times\dfrac{3\times10^{8}}{500\times10^{-9}}

E_{2}=4.002\times10^{-19}\ J

So, E₂>E₁

Hence, The maximum kinetic energy of the electrons emitted from the surface will be greater.

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Answer:

65 m/s

Explanation:

v=v0+at <=> v = 11 + 12 t ∧ t = 4.5 s <=> v = 11 + 12×4.5 <=> v = 65 m/s

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Nastasia [14]

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When a sinusoidal wave with speed 20 m/s , wavelength 35 cm and amplitude of 1.0 cm passes, what is the maximum speed of a point
vova2212 [387]

To solve this problem it is necessary to apply the concepts related to frequency as a function of speed and wavelength as well as the kinematic equations of simple harmonic motion

From the definition we know that the frequency can be expressed as

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Where,

v = Velocity \rightarrow 20m/s

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Therefore the frequency would be given as

f = \frac{20}{35*10^{-2}}

f = 57.14Hz

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\omega = 2\pi *57.14

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Now the maximum speed from the simple harmonic movement is given by

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