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Leno4ka [110]
3 years ago
6

Consider a laser pointer that emits red light with wavelength 650 nm. This light is used for a photoelectric effect experiment w

here the anode in the evacuated glass tube is made up of a material that has work function equal to 1 eV. 1. What is the energy of an individual photon that comes out of the laser pointer?
Physics
1 answer:
Vikentia [17]3 years ago
4 0

Answer:

<em>The energy of an individual photon that comes out of the laser pointer is 1.91 eV</em>

Explanation:

The energy of a photon can be obtained using the expression below

E = hc/λ

where E is the energy;

h is the Planck's constant  = 6.626 x 10-34 Js;

c is the speed of light = 3.00 x 10^{8} m/s (speed of light);

λ is the wavelength = 650 nm =650 x 10^{-9} m.

E = (6.626 x 10-34 Js) x (3.00 x 10^{8} m/s) /650 x 10^{-9} m

E = 3.058 x 10^{-19} J

1 joule = 6.242 x 10^{18} eV

3.058 x 10^{-19} J = 3.058 x 10^{-19} J x 6.242 x 10^{18} eV = 1.91 eV

Therefore the energy of an individual photon that comes out of the laser pointer is 1.91 eV

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Answer:

5 ohms

Explanation:

Given:

EMF of the ideal battery (E) = 60 V

Voltage across the terminals of the battery (V) = 40 V

Current across the terminals (I) = 4 A

Let the internal resistance be 'r'.

Now, we know that, the voltage drop in the battery is given as:

V_d=Ir

Therefore, the voltage across the terminals of the battery is given as:

V= E-V_d\\\\V=E-Ir

Now, rewriting in terms of 'r', we get:

Ir=E-V\\\\r=\frac{E-V}{I}

Plug in the given values and solve for 'r'. This gives,

r=\frac{60-40}{4}\\\\r=\frac{20}{4}\\\\r=5\ ohms

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4 years ago
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Answer:

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Explanation:

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