Physics: When light with a frequency f1 = 547.5 thz illuminates a metal surface, the most energetic photoelectrons have 1.260 x 10^-19 j of kinetic energy. when light with a frequency f2 = 738.8 thz is used instead, the most energetic photo-electrons have 2.480 x 10^-19 j of kinetic energy using these experimental results, determine the approximate value of planck s constant.

When light with a frequency f1 = 547.5 thz illuminates

Ответ:

ajay67

11.01.2022

Physics

The approximate value of Planck's constant is $6.377\times10^{-34}\ J$

Explanation:

Given that,

Frequency $f_{1}= 547.5\ THz$

Kinetic energy $K.E=1.260\times10^{-19}\ J$

Frequency $f_{2}=738.8\ THz$

Kinetic energy $K.E=2.480\times10^{-19}\ J$

We need to calculate the approximate value of Planck's constant

Using formula of change in energy

E = hf

$K.E_{2}-K.E_{1}=h(f_{2}-f_{1})$

$h=\dfrac{K.E_{2}-K.E_{1}}{(f_{2}-f_{1})}$

$h=\dfrac{2.480\times10^{-19}-1.260\times10^{-19}}{738.8\times10^{12}-547.5\times10^{12}}$

$h=6.377\times10^{-34}\ J$

Hence, The approximate value of Planck's constant is $6.377\times10^{-34}\ J$

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Ответ:

jazzy9927

21.10.2020

Physics

1. The resulting image is formed at 2F

2. The image formed is real and inverted

3. Magnification is 1

Explanation:

1. The image is formed at 2F

When an object is placed at point 2F of a convex lens, the resulting image will also be located at the point 2F on the other side of the lens.

This to say object length is equal to image length.

Using Equations;

u = object distance from the mirror = 2F

v = image distance from the mirror

f = focal length

$\frac{1}{u} -\frac{1}{v} =\frac{1}{f}$

substitute 2F for u

$\frac{1}{2f} -\frac{1}{v} =\frac{1}{f}$

Therefore $\frac{1}{2f} -\frac{1}{f} =\frac{1}{v}$

$\frac{-1}{2f} =\frac{1}{v}$

And v = -2f

i.e Image is formed at 2F

2. The image formed is real and inverted;

Real image means the image can be placed on a screen, i.e. light rays will converge at a location which exists in reality and this is accompanied by an inversion whereby an upright object forms an image that is upside-down.

3. Magnification is 1;

Magnification of a convex lens is the ratio of an object height to image height OR the ratio of distance of object to distance of image i.e. $\frac{u}{v}$

where u = 2f

and v = 2f

M = 1

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