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kupik [55]
3 years ago
10

The light intensity incident on a metallic surface with a work function of 3 eV produces photoelectrons with a maximum kinetic e

nergy of 2 eV. The frequency of the light is doubled. Determine the maximum kinetic energy (in eV).
Physics
1 answer:
Vlad1618 [11]3 years ago
7 0

Answer:

7ev

Explanation:

The Energy of an incident light is given as E = hf;

Where h is Planks constant

f is frequency

Also.

E = Eo + K.E

Where Eo is work function and K.E is kinetic energy.

Hence substituting the given values,

E = 3ev + 2ev = 5ev

The frequency of the incident wave if it's doubled the Energy would be doubled since E is proportional to f with other elements been constant.

Hence :E = 10ev

But K.E = E -Eo

= 10ev - 3ev = 7ev

Maximum Kinetic Energy is 7ev.

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3

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because

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In which ancient civilization did the sport of cucj originate
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Which of the following planets rotates in the opposite direction than the other planets? Venus Neptune Saturn Uranus
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The answer is option A "Venus".

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The planet Venus spins the wrong way many scientists are not sure why. Its not options B, C, or D because these planets spin the same way the as each other. (besides Venus) Therefore the answer is option A.

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In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this
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Answer:

1). v = - 2960526m/s

2). Toward us

3). v = - 493421m/s

4). Toward us

5). v = 1480263m/s

6).  Away from us

7). v = 3207236m/s

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm (\lambda_{0} = 121.6nm)

Redshift: \lambda_{measured} > \lambda_{0}

Blueshift: \lambda_{measured} < \lambda_{0}

Then, for this particular case it is gotten:

Star 1: \lambda_{measured} = 120.4nm

Star 2: \lambda_{measured} = 121.4nm

Star 3: \lambda_{measured} = 122.2nm

Star 4: \lambda_{measured} = 122.9nm

Star 1:

Blueshift: 120.4nm < 121.6nm

Toward us

Star 2:

Blueshift: 121.4nm < 121.6nm

Toward us

Star 3:

Redshift: 122.2nm > 121.6nm

Away from us

Star 4:

Redshift: 122.9nm > 121.6nm

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

v = c\frac{\Delta \lambda}{\lambda_{0}}  (1)

Where \Delta \lambda is the wavelength shift, \lambda_{0} is the wavelength at rest, v is the velocity of the source and c is the speed of light.

v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}}) (2)

<em>Case for star 1 \lambda_{measured} = 120.4 nm:</em>

<em></em>

v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})

v = - 2960526m/s

Notice that the negative velocity means that is approaching to the observer.

<em>Case for star 2 \lambda_{measured} = 121.4 nm:</em>

v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})

v = - 493421m/s

<em>Case for star 3 \lambda_{measured} = 122.2 nm:</em>

v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})

v = 1480263m/s

<em>Case for star 4 \lambda_{measured} = 122.9 nm:</em>

v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})

v = 3207236m/s

4 0
3 years ago
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