Question

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

Answer 1

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.

Answer 2

We have that from the Question, it can be said that  the maximum kinetic energy of the electrons emitted from the surface will be Greater

Greater

Option C

From the Question we are told

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

Generally the equation for Energy is mathematically given as

E=hf

Therefore

the equation for frequency is mathematically given as

$v=\lambda f$

Therefore

with increasing frequency there is decreasing  wavelength

Hence

the maximum kinetic energy of the electrons emitted from the surface will be Greater

Greater

Option C

For more information on this visit

https://brainly.com/question/19007362?referrer=searchResults