Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
Answer:
The image distance is 17.56 cm
Explanation:
We have,
Height of light bulb is 3 cm.
The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm
Focal length of the lens, f = +13 cm
Let v is distance between image and the lens. Using lens formula :

So, the image distance is 17.56 cm.
5-ohm
Extra
Variable
120-ohm
Variable
Pg. 614
Answer:
v = 3200 m/s
Explanation:
As we know that the frequency of the sound wave is given as

wavelength of the sound wave is given as

so now we have

so we will have

