Download "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/"
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FurD88UmnrkQ%252Fsddefault.jpg&w=1139&q=75)
Similar videos from our catalog
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252F6ohCc_A7bUw%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FBq8CQgOvaQU%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FEdNjO5qj7Zk%252Fsddefault.jpg&w=1139&q=75)
![Обучение грамоте. 1 класс. Звук [Ы] и буква Ы /27.10.2020/](/_next/image?url=https%3A%2F%2Funidownloader.com%2FcontentImages%2FXrBWfaw1P1fr8eQS8-hxwI3xf_M%3D%2Ffilters%3Aquality(80)%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FMox5McztrhU%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FO7WM0fXKMJI%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FQC_83YUI8J0%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FhUG7scqwwRI%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FqZW-iHlTHXE%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252Fr6CsU_u2exg%252Fsddefault.jpg&w=1139&q=75)
%3Aformat(webp)%2Fi.ytimg.com%252Fvi%252FrKAGekLhtFs%252Fsddefault.jpg&w=1139&q=75)
Video tags
Subtitles
Russian
00:00:02
[music]00:00:05
[applause]00:00:07
[music]00:00:14
hello topic of the lesson solving problems on00:00:18
topics atomic physics00:00:21
quantum physics and physics of the atomic nucleus00:00:28
Today in the lesson you will learn to apply the00:00:31
Stefan-Boltzmann law to describe the00:00:35
thermal radiation of a black00:00:38
body,00:00:39
use the Einstein equation for the00:00:41
photoelectric effect, the wavelength formula de00:00:45
broil and the formula for00:00:47
radioactive decay when solving the00:00:50
problem also calculate the binding energy of00:00:54
the atomic nucleus specific binding energy when00:00:59
solving problems we will solve the problem problem number00:01:03
one conditions determine00:01:06
the energy emitted in five minutes and a00:01:09
viewing window with an area of 8 centimeters00:01:13
square melting furnace considering it an00:01:17
absolutely black body the temperature of the furnace00:01:20
is 1000 kelvin the00:01:24
Stefan-Boltzmann constant sigma00:01:27
is 5 point 67 multiplied by ten00:01:32
to the minus eighth power in from divided00:01:36
by the product of meter squared and00:01:39
kelvin to the fourth power let's start00:01:42
solving the problem00:01:48
yes for t 5 minutes with 8 centimeters00:01:54
squared t00:01:56
absolute temperature 1000 kelvin00:01:59
sigma00:02:00
we spelled it out in the statement of the problem, we00:02:04
need to find the energy of the emitted00:02:08
body, we use the system 75 minutes is 30000:02:15
seconds and00:02:16
with 8 10 meters squared this is 8 multiplied by00:02:22
10 to the minus fourth power of meters00:02:25
squared let00:02:26
's proceed to solving the problem p with index00:02:30
t that is, the energetic luminosity00:02:33
is calculated according to the formula,00:02:35
energy is divided by the product x00:02:40
multiplied by t,00:02:42
then also according to the Stefan-Boltzmann law, it00:02:44
is known that this is equal to the product of00:02:48
sigma with t large to the fourth00:02:51
power, we equate these formulas from00:02:56
this relationship, we find what00:03:00
the energy is equal to, it will be calculated using this00:03:03
formula, we will substitute the value, we will perform00:03:07
calculations we find that the energy00:03:10
is 13000 608 joules we convert to00:03:16
kilojoules the energy is 13.6 body00:03:21
joules we write the answer to the problem the00:03:24
energy is 13.6 kilojoules00:03:30
the problem is solved problem number 200:03:36
determine the work function of the electron from the00:03:40
surface of the photocathode00:03:42
and the red boundary of the photoelectric effect if when the00:03:47
photocell is irradiated with light frequency00:03:51
1.66 degrees herz00:03:55
photo current stops when the blocking00:03:58
voltage is 4.100:04:00
volts, we write the condition of the problem y the00:04:10
blocking voltage is 4.1 volts not 1.600:04:16
multiplied by 10 15 degrees hertz charge of the00:04:20
electron is one integer 60th by 10 minus00:04:23
19 steppe constant bar 6 point00:04:27
626 thousandths multiplied by 10 minus the00:04:30
thirty-fourth power jul00:04:33
multiplied by a second we need to find00:04:35
the work function and the red limit of the00:04:38
photoelectric effect, that is, not the minimum, let's00:04:43
start solving the problem, we know that00:04:47
when locking, the voltage energy and00:04:51
multiplied by z will be equal to the kinetic00:04:56
energy m in square divided by two,00:05:00
we also know the Einstein equation for00:05:03
the photoelectric effect, instead of the kinetic energy,00:05:07
we substitute00:05:09
the product e y in the science set,00:05:14
we get the following equation as much as e00:05:18
multiplied by z plus the work function00:05:22
from here the work function is equal to the difference between a00:05:28
lot and the blocking voltage,00:05:32
we also know that the red boundary00:05:35
not y minimum this will be calculated00:05:42
using the formula a of the output divided by the00:05:45
plank constant00:05:51
substituting the values the formula for the work function we will00:05:54
perform the calculations we will get that the00:05:59
work function is approximately00:06:02
4 times 10 minus 19 powers of00:06:05
joules we will also calculate the red limit00:06:10
we will substitute the values we will perform the calculation00:06:13
not y minimum is 6 times at00:06:18
10 14 powers of hertz we write the answer to the problem, the00:06:22
work function is approximately00:06:25
4 times 10 minus 19 powers of00:06:28
joules and not minimal, that is, the red00:06:32
limit of the photoelectric effect is 6 times00:06:35
10 14 powers of hertz solved the00:06:42
third problem the electron00:06:46
whose initial speed can be neglected00:06:48
passed the accelerating potential difference00:06:52
51 volt determine the de00:06:57
Broglie wavelength the mass of the electron is 9.1 to 10 minus00:07:02
31 degrees kilogram and the charge of the electron is00:07:05
one integer 60's to 10 minus 19 degrees00:07:09
coulomb we write the condition of the problem at 51 volts the00:07:19
mass of the electron is nine point one00:07:22
to 10 minus 31 degrees kilogram00:07:25
charge electron 1.69 to that power00:07:30
coulomb Planck's constant 6 point00:07:33
626 thousandths times 10 minus 34 to the00:07:38
power of joule per second we need to00:07:41
find00:07:42
lambda we can use the formula to00:07:48
calculate the de Broglie wavelength00:07:52
this is the ratio of Planck's constant to the00:07:57
product of m times v00:08:01
also we we know that multiply by y is equal to00:08:06
m squared divide by two from this00:08:11
ratio we will find what is equal to in00:08:15
then we will substitute the values in the first00:08:18
formula and carry out the transformations00:08:21
we will get that lambda will be equal to the ratio of the00:08:25
constant plank by the square root00:08:27
of 2 m multiplied by u we will00:08:36
set aside the values we will find what is the00:08:39
de Broglie wavelength for an00:08:42
electron Linda will be approximately00:08:46
1.75 times 10 to the minus00:08:51
tenth power of meters we also know00:08:54
that 1 pica meter is 10 to the minus 12th00:08:57
power of meters lambda will be00:09:00
approximately one hundred seventy00:09:02
two pica meters let's write down the answer to the problem the00:09:06
de Broglie wavelength is00:09:09
approximately00:09:11
172 and k meters the problem is solved00:09:18
the following problem 400:09:20
some radioactive isotope has a00:09:24
half-life you what part of I der00:09:28
decayed in time 3t00:09:31
write yes at t equals 3 t large00:09:38
that is, with three half-lives we00:09:41
need to find delta n let's start00:09:45
solving the problem delta n and that is, the00:09:48
difference n is zero, that is, the00:09:51
initial number of nuclei minus n00:09:57
according to the law of radioactive decay00:10:00
n is equal to n 0 multiplied by 2 to the power00:10:05
minus t small divided by t00:10:08
large that is, the half-life00:10:14
we will find the ratio ncr lion and also00:10:19
substitute instead of t small three00:10:24
half-lives the00:10:26
half-life is reduced we00:10:29
get that the ratio will be equal to 2 to the00:10:34
power of minus 3 or this ratio00:10:40
will be equal to 1 8 that is, n will be equal to 100:10:46
8 of the00:10:48
original number leader from we00:10:53
will assume that 0 is equal to one00:10:57
then taking into account the fact that n and that is one00:11:01
eighth and zero we substitute and00:11:06
delta n will be equal to seven eighths00:11:10
n 0 or the decimal fraction00:11:14
it will be 0 point 875 thousandths and 000:11:20
we will write the answer to the problem00:11:23
delta n is zero point00:11:29
875 thousandths and 000:11:33
the problem is solved00:11:37
we will solve fifth problem, calculate the specific00:11:41
binding energy of nucleons in an iron nucleus with an00:11:45
atomic mass of 56 and an ordinal number of 26,00:11:51
we write down the condition of the problem:00:11:57
iron with a mass number of 56 and an ordinal00:12:01
number of 26 means a will be equal to 56 z 2600:12:10
the mass of a proton is 1 point 72800:12:19
hundred thousandths of an atomic mass unit the mass of a00:12:22
neutron is 1 point 86600:12:28
hundred thousandths of atomic mass units and the mass of the00:12:32
iron nucleus is 55 point 90 349400:12:42
hundred thousandths of atomic mass units00:12:45
find the specific binding energy let's00:12:49
start solving the problem specific binding00:12:52
energy is00:12:53
equal to the ratio of the binding energy to the00:12:56
mass number00:13:01
we need to find the number of neutrons00:13:04
n equals a minus z and it will be00:13:11
30 neutrons the00:13:15
binding energy is calculated by the formula00:13:18
delta m multiplied by 931.5 mega00:13:24
electron volts delta m is a mass defect00:13:28
it is calculated using this formula let's00:13:31
substitute the values let's do the calculations00:13:35
delta m defect most is approximately00:13:38
zero point00:13:42
514 thousandths of atomic mass units we will00:13:47
further calculate with mom what is the00:13:51
binding energy equal? Let's substitute the values Let's do00:13:54
the calculations The binding energy00:13:57
is approximately four hundred00:14:00
seventy-eight point nine 121400:14:05
ten-thousandths mega electron-volts Let00:14:08
's find the specific binding energy Let's substitute the00:14:12
values Let's also00:14:14
do the calculation The specific energy00:14:18
is 8 point fifty-five hundredths00:14:23
mega electron-volts We'll write down the00:14:28
answer problem the specific energy is 800:14:33
point fifty-five hundredths00:14:35
mega electron-volts per slope, let's00:14:39
summarize the lesson today in the lesson we learned00:14:44
to apply the Stefan-Boltzmann law00:14:46
to describe the thermal radiation of a00:14:49
black body, use the00:14:53
Einstein equation for the photoelectric effect, the00:14:56
de Broglie wavelength formula and the formula for the00:15:00
law of radioactive decay at00:15:03
solving problems also calculate the00:15:07
binding energy and specific binding energy of the atomic00:15:12
nucleus when solving problems lesson is over00:15:19
goodbye00:15:23
[music]00:15:26
[applause]00:15:28
[music]Description:
Тема урока: Урок решения задач по темам: «Атомная физика». «Квантовая физика». «Физика атомного ядра» С понедельника по пятницу с 09:00-18:00 телевизионные уроки в эфире EL ARNA Запись телеуроков: https://www.youtube.com/playlist?list=PLDIwNJYcIQ77fZ7Tqrb5myYpNtQbhnA4p https://edu.elarna.kz/ru/tv-lessons
Preparing download options
* — If the video is playing in a new tab, go to it, then right-click on the video and select "Save video as..."
** — Link intended for online playback in specialized players
Questions about downloading video
How can I download "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/" video?
http://unidownloader.com/ website is the best way to download a video or a separate audio track if you want to do without installing programs and extensions.
The UDL Helper extension is a convenient button that is seamlessly integrated into YouTube, Instagram and OK.ru sites for fast content download.
UDL Client program (for Windows) is the most powerful solution that supports more than 900 websites, social networks and video hosting sites, as well as any video quality that is available in the source.
UDL Lite is a really convenient way to access a website from your mobile device. With its help, you can easily download videos directly to your smartphone.
Which format of "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/" video should I choose?
The best quality formats are FullHD (1080p), 2K (1440p), 4K (2160p) and 8K (4320p). The higher the resolution of your screen, the higher the video quality should be. However, there are other factors to consider: download speed, amount of free space, and device performance during playback.
Why does my computer freeze when loading a "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/" video?
The browser/computer should not freeze completely! If this happens, please report it with a link to the video. Sometimes videos cannot be downloaded directly in a suitable format, so we have added the ability to convert the file to the desired format. In some cases, this process may actively use computer resources.
How can I download "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/" video to my phone?
You can download a video to your smartphone using the website or the PWA application UDL Lite. It is also possible to send a download link via QR code using the UDL Helper extension.
How can I download an audio track (music) to MP3 "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/"?
The most convenient way is to use the UDL Client program, which supports converting video to MP3 format. In some cases, MP3 can also be downloaded through the UDL Helper extension.
How can I save a frame from a video "Физика. 11 класс. Урок решения задач по темам: «Атомная физика». «Квантовая физика» /24.05.2021/"?
This feature is available in the UDL Helper extension. Make sure that "Show the video snapshot button" is checked in the settings. A camera icon should appear in the lower right corner of the player to the left of the "Settings" icon. When you click on it, the current frame from the video will be saved to your computer in JPEG format.
What's the price of all this stuff?
It costs nothing. Our services are absolutely free for all users. There are no PRO subscriptions, no restrictions on the number or maximum length of downloaded videos.