Love from india... sir ur explaination is super amazing... im so surprised... ur lectures ❤... pls make such vid.. we all here always!!❤ here class 11 student ThankThankyou!!
You forgot to multiply to the 10^-34 the 10^14 from the frequency value. Thus, the result would be E = 4.97 * 10^-19J. Also, 6.62 * 7.5 is 49.65 not 4.96.
He had it correct with the 4.96. He just changed the base 10 from 49.65 x 10^-20 to 4.96 x 10^-19. When he wrote the final solution he just wrote the wrong superscript.
Why did you round to 3 significant figures for the final problem? Isn't the value with the least amount of significant figures the wavelength of 400 nm (only 1 SigFig)?
You are correct. I should have rounded to one significant figure since 400 nm has only one significant figures. One of my weak points is that I am sometimes inconsistent in my use of significant figures. Interesting, textbooks are also sometimes inconsistent in the use of significant figures. I should have listed the wavelength as 4.00 x 10^2 nm, and then I could have rounded to three significant figures.
Sir , I'm from India....and I feel that American highschool 11 & 12 grades textbooks would be helpful for me to learn.....so it would be great if you can send me the name of those textbooks...and I'll find ways to buy it here....thank you
One really old textbook from 1998 that I like to use is Merrill Chemistry written by Robert Smoot. The ISBN number is 0-02-825527-5. Another book I've used that is more recent (2002) is Chemistry: Matter and Change. The ISBN for this book is 0-02-828379-1.
I don't know why I have not come across this channel b4. this is the best channel so far. thank you so much
Love from india... sir ur explaination is super amazing... im so surprised... ur lectures ❤... pls make such vid.. we all here always!!❤ here class 11 student ThankThankyou!!
there is no other video better than this. I am so bad at chem and if I can understand it I am 100% sure everyone else can. Thanks a lot.
I love your work u make it simple and understandable i’m going to right my external exam some years from now and your really helping ❤❤❤❤❤❤❤
#im🇳🇬Nigerian
Amazing lectures,sir.absolutely amazing...I completely understood the topic.
You forgot to multiply to the 10^-34 the 10^14 from the frequency value. Thus, the result would be E = 4.97 * 10^-19J. Also, 6.62 * 7.5 is 49.65 not 4.96.
Sorry for the mistakes! Thanks for the corrections.
He had it correct with the 4.96. He just changed the base 10 from 49.65 x 10^-20 to 4.96 x 10^-19. When he wrote the final solution he just wrote the wrong superscript.
I love your enthusiasm!
Well understood 🤝
Amazing lectures series.
Thank you sir 🤗
Very interesting
U r the best
I need to know what kind of pen he is using!
I'm using a black, Papermate Flair pen (medium point).
Thank you so much much sirrrr😊😊😊😊😊😊😊😊
Why did you round to 3 significant figures for the final problem? Isn't the value with the least amount of significant figures the wavelength of 400 nm (only 1 SigFig)?
You are correct. I should have rounded to one significant figure since 400 nm has only one significant figures. One of my weak points is that I am sometimes inconsistent in my use of significant figures. Interesting, textbooks are also sometimes inconsistent in the use of significant figures. I should have listed the wavelength as 4.00 x 10^2 nm, and then I could have rounded to three significant figures.
Sounds good, thanks for the reply!
Sir , I'm from India....and I feel that American highschool 11 & 12 grades textbooks would be helpful for me to learn.....so it would be great if you can send me the name of those textbooks...and I'll find ways to buy it here....thank you
One really old textbook from 1998 that I like to use is Merrill Chemistry written by Robert Smoot. The ISBN number is 0-02-825527-5. Another book I've used that is more recent (2002) is Chemistry: Matter and Change. The ISBN for this book is 0-02-828379-1.
Thanks...😊☺😊