When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10-19 J. Determine the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

Answers

Answer 1

Answer:

The new wavelength is 112.5 nm.

Explanation:

It is given that,

When light with a wavelength of 225 nm is incident on a certain metal surface, electrons are ejected with a maximum kinetic energy of 2.98 × 10⁻¹⁹ J. We need to find the wavelength (in nm) of light that should be used to double the maximum kinetic energy of the electrons ejected from this surface.

The energy of incident electron is given by :

[tex]E=\dfrac{hc}{\lambda}[/tex]

New energy is 2 E and new wavelength is [tex]\lambda'[/tex]. So,

[tex]\dfrac{E}{2E}=\dfrac{\lambda'}{\lambda}\\\\\dfrac{1}{2}=\dfrac{\lambda'}{\lambda}\\\\\lambda'=\dfrac{\lambda}{2}\\\\\lambda'=\dfrac{225}{2}\\\\\lambda=112.5\ nm[/tex]

So, the new wavelength is 112.5 nm.


Related Questions

Please help in the 2nd question

Answers

Answer:

[tex]q=4\times 10^{-16}\ C[/tex]

Explanation:

It is given that,

The charge on an object is 2500 e.

We need to find how many coulombs in the object. The charge remains quantized. It says that :

q = ne

[tex]q=2500\times 1.6\times 10^{-19}\ C\\\\q=4\times 10^{-16}\ C[/tex]

So, the charge on the object is [tex]4\times 10^{-16}\ C[/tex].

Four heavy elements (A, B, C, and D) will fission when bombarded by neutrons. In addition to fissioning into two smaller elements, A also gives off a beta particle, B gives off gamma rays, C gives off neutrons, and D gives off alpha particles. Which element would make a possible fuel for a nuclear reactor

Answers

Answer:

Element C will be best for a nuclear fission reaction

Explanation:

Nuclear fission is the splitting of the nucleus of a heavy atom by bombarding it with a nuclear particle. The reaction leads to the the atom splitting into two smaller elements and a huge amount of energy is liberated in the process. For the reaction to be continuous in a chain reaction, the best choice of element to use as fuel for the reaction should be the element whose nucleus also liberates a neutron particle after fission. The neutron that is given off by other atoms in the reaction will then proceed to bombard other atoms of the element in the reaction, creating a cascade of fission and bombardment within the nuclear reactor.

How far does a roller coaster travel if it accelerates at 2.83 m/s2 from an initial
velocity of 3.19 m/s for 12.0 s?

Answers

Answer:

b

Explanation:

A 60-kg skier is stationary at the top of a hill. She then pushes off and heads down the hill with an initial speed of 4.0 m/s. Air resistance and the friction between the skis and the snow are both negligible. How fast will she be moving after she is at the bottom of the hill, which is 10 m in elevation lower than the hilltop

Answers

Answer:

The velocity is  [tex]v = 8.85 m/s[/tex]

Explanation:

From the question we are told that

    The mass of the skier is [tex]m_s = 60 \ kg[/tex]

      The initial speed is [tex]u = 4.0 \ m/s[/tex]

       The height is  [tex]h = 10 \ m[/tex]

According to the law of energy conservation

     [tex]PE_t + KE_t = KE_b + PE_b[/tex]

Where [tex]PE_t[/tex] is the potential energy at the top which is mathematically evaluated as

       [tex]PE_t = mg h[/tex]

substituting values

       [tex]PE_t = 60 * 4*9.8[/tex]

      [tex]PE_t = 2352 \ J[/tex]

And  [tex]KE_t[/tex] is the kinetic energy at the top which equal to zero due to the fact that velocity is zero at the top of the hill

And  [tex]KE_b[/tex] is the kinetic energy at the bottom of the hill which is mathematically represented as

         [tex]KE_b = 0.5 * m * v^2[/tex]

  substituting  values

         [tex]KE_b = 0.5 * 60 * v^2[/tex]

=>     [tex]KE_b = 30 v^2[/tex]

Where v is the velocity at the bottom

   And [tex]PE_b[/tex] is the potential  energy at the bottom which equal to zero due to the fact that height  is zero at the bottom of the hill

So  

        [tex]30 v^2 = 2352[/tex]

=>      [tex]v^2 = \frac{2352}{30}[/tex]

=>       [tex]v = \sqrt{ \frac{2352}{30}}[/tex]

        [tex]v = 8.85 m/s[/tex]

         

Answer:

The Skier's velocity at the bottom of the hill will be 18m/s

Explanation:

This is simply the case of energy conversion between potential and kinetic energy. Her potential energy at the top of the hill gets converted to the kinetic energy she experiences at the bottom.

That is

[tex]mgh = 0.5 mv^{2}[/tex]

solving for velocity, we will have

[tex]v= \sqrt{2gh}[/tex]

hence her velocity will be

[tex]v=\sqrt{2 \times 9.81 \times 10}=14.00m/s[/tex]

This is the velocity she gains from the slope.

Recall that she already has an initial velocity of 4m/s. It is important to note that since velocities are vector quantities, they can easily be added algebraically. Hence, her velocity at the bottom of the hill is 4 + 14 = 18m/s

The Skier's velocity at the bottom of the hill will be 18m/s

What is the gravitational potential energy of a ball of mass 2.00 kg which is tossed to a height of 13.0 m above the ground? Answer in J, taking the potential energy to be 0.00 J at the ground.

Answers

Answer:

I believe the answer is 254.8 J, or rounded 255 J.

Explanation:

The formula for potential energy is:

PE=m(h)g

This means the mass (m) times height (h) times gravity (m). Gravity is 9.8 m/s (meters per second). Putting all of the numbers into it would equal:

PE=2(13)9.8

This equals 254.8 exactly, or if the assignment calls for you to round, 255.

A body moves due north with velocity 40 m/s. A force is applied

on it and the body continues to move due north with velocity 35 m/s. W. .What is the direction of rate of change of momentum,if it takes

some time for that change and what is the direction of applied

external force?​

Answers

Answer:

the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

Explanation:

The change in the momentum of a body, if the mass of the body is constant, is given by the following formula:

[tex]\Delta p=\Delta (mv)\\\\\Delta p=m\Delta v[/tex]

p: momentum

m: mass

[tex]\Delta v[/tex]:  change in the velocity

The sign of the change in the velocity determines the direction of rate of change. Then you have:

[tex]\Delta v=v_2-v_1[/tex]

v2: final velocity = 35m/s

v1: initial velocity = 40m/s

[tex]\Delta v =35m/s-40m/s=-5m/s[/tex]

Hence, the direction of rate of change of the momentum is against the motion of the body, that is, downward.

The applied force is also against the direction of motion of the body, downward.

You hold block A with a mass of 1 kg in one hand and block B with a mass of 2 kg in the other. You release them both from the same height above the ground at the same time. (Air resistance can safely be ignored as they fall.) Which of the following describes and explains the motion of the two objects after you release them?
A) Before A hits the ground before the block B. The same force of gravity acts on both objects, but because block A has a lower mass, it will have a larger acceleration
B) Block B hits the ground before the block A. The larger force on block B causes it to have a greater acceleration than block A
C) The two blocks hit the ground at the same time, because they both ackelerate at the same rate while falling. The greater gravitational force on block B is compersated giving it the same acceleration
D) The two blocks hit the ground at the same time because they move at the same constant speed while falling.The force on the blocks are irrelevent once they start falling

Answers

ok so im not that good when it comes to physics but i think its C)

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