The values of Planck's constant and the speed | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The relationship between wavelength $(\lambda)$,speed of light $(c)$,and frequency $(
u)$ is given by $\lambda = \frac{c}{
u}$.Given: $c = 3.0 \

Vedclass · Accepted Answer

$4 	imes 10^1$

Vedclass · Answer

(D) The relationship between wavelength $(\lambda)$,speed of light $(c)$,and frequency $(
u)$ is given by $\lambda = \frac{c}{
u}$.Given: $c = 3.0 	imes 10^8 \ m \ s^{-1}$ and $
u = 8 	imes 10^{15} \ s^{-1}$.Substituting the values: $\lambda = \frac{3.0 	imes 10^8}{8 	imes 10^{15}} = 0.375 	imes 10^{-7} \ m = 3.75 	imes 10^{-8} \ m$.To convert meters to nanometers $(nm)$,multiply by $10^9$: $\lambda = 3.75 	imes 10^{-8} 	imes 10^9 \ nm = 37.5 \ nm$.The approximate value is $4 	imes 10^1 \ nm$.

The values of Planck's constant and the speed of light are $6.63 \times 10^{-34} \ J \ s$ and $3.0 \times 10^8 \ m \ s^{-1}$ respectively. Find the approximate wavelength in nanometers $(nm)$ of a light ray with a frequency of $8 \times 10^{15} \ s^{-1}$.

Similar Questions

In Bohr's model of the atom,when an electron jumps from $n=1$ to $n=3$,how much energy will be absorbed?

In which hydrogen-like ion is the difference in wavelength between the first line of the Balmer series and the first line of the Lyman series equal to $59.3 \ nm$? $[R_H = 109678 \ cm^{-1}]$

Circumference of the first orbit of hydrogen atom is given by the formula :

Based on equation $E = -2.178 \times 10^{-18} \ J \ \frac{Z^2}{n^2}$,certain conclusions are written. Which of them is not correct?

Assertion : Angular momentum of an electron in any orbit is given by angular momentum $= \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Reason : The principal quantum number,$n$,can have any integral value.

Vedclass Test Series

Exam Paper Generator

Online Exam Module