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3 years ago

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Upon being struck by 240-nm photons, a metal ejects electrons with a maximum kinetic energy of 1.45 eV. What is the work functio

n of this metal? (A) 3.73 eV (B) 3.13 eV (C) 4.33 eV (D) 4.92 eV

1 answer:

Bess [88]3 years ago

3 0

Answer:

W = 3.73 eV

Explanation:

Given data:

wavelength of photons 240 nm

kinetic energy 1.45 eV

B y using photoelectric equation

Kinetic Energy = Photon Energy - work friction

$K.E = \frac{hc}{\lambda} -W$

$\frac{hc}{\lambda} = \frac{6.63\times 10^{-34}\times 3\times 10^8}{240\times 10^{-9}\times 1.6\times 10^{-19}} eV$

= 5.179 eV

= 5.18 eV $W = \frac{hc}{\lambda} - K.E$

= (5.18 - 1.45) eV

W = 3.73 eV

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Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

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And substituting into (1),

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We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

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$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

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Substituting these values into the equation, we find the time of flight of the shell:

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3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

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Therefore, the final speed of the shell is:

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Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

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