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nlexa [21]
3 years ago
9

A metal has a work function of 4.50 eV. Find the maximum kinetic energy of the photoelectrons if light of wavelength 250 nm shin

es on the metal. (1 eV = 1.60 × 10−19 J, c = 3.00 × 108 m/s,h = 6.626 × 10-34 J ∙ s)
A) 0.00 eV.
B) 0.37 eV.
C) 0.47 eV.
D) 0.53 eV.
Physics
1 answer:
VladimirAG [237]3 years ago
5 0

Answer:

0.46 eV

Explanation:

The maximum kinetic energy of the photoelectrons can be found using the formula

K.E = h(c)/λ - h(c)/λ(T), where

K.E = maximum kinetic energy needed

h(c) = Planck's constant, 6.626*10^-34

λ = Wavelength of the light, 250 nm

To start with, we convert our Planck's constant to nm and we have

h(c) = 1240 eV nm

Again, from the question, we already have h(c)/λ(T) given as 4.5 eV so all we need to do is substitute for it.

Plugging in all our values, we have

K.E = 1240/250 - 4.5

K.E = 4.96 - 4.5

K.E = 0.46 eV

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3 years ago
When is the velocity of a mass on a spring at its maximum value?
ehidna [41]

Answer:

A.  when the mass has a displacement of zero

Explanation:

The velocity of a mass on a spring can be calculated by using the law of conservation of energy. In fact, the total energy of the mass-spring system is equal to the sum of the elastic potential energy (U) of the spring and the kinetic energy (K) of the mass:

E=U+K=\frac{1}{2}kx^2 + \frac{1}{2}mv^2

where

k is the spring constant

x is the displacement of the mass with respect to the equilibrium position of the spring

m is the mass

v is the velocity of the mass

Since the total energy E must remain constant, we can notice the following:

- When the displacement is zero (x=0), the velocity must be maximum, because U=0 so K is maximum

- When the displacement is maximum, the velocity must be minimum (zero), because U is maximum and K=0

Based on these observations, we can conclude that the velocity of the mass is at its maximum value when the displacement is zero, so the correct option is A.


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4 years ago
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A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
KATRIN_1 [288]

Complete Question

Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU.  What is its orbital period?

Answer:

The value  is  T_R  = 11.8 \  days  

Explanation:

From the question we are told that

   The semi - major axis of the rocky debris  a_R = 45.0\  AU

   The semi - major axis of  Planet D is  a_D  = 60 \  AU

    The orbital  period of planet D is  T_D = 18.164 \  days

Generally from Kepler third law

          T \  \ \alpha \ \ a^{\frac{3}{2} }

Here T is the  orbital period  while a is the semi major axis

So  

        \frac{T_D}{T_R}  =  \frac{a^{\frac{3}{2} }}{a_R^{\frac{3}{2} }}

=>     T_R  = T_D *  [\frac{a_R}{a_D} ]^{\frac{3}{2} }  

=>     T_R  = 18.164  *  [\frac{ 45}{60} ]^{\frac{3}{2} }

=>      T_R  = 11.8 \  days  

   

7 0
3 years ago
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