A source of monochromatic radiation of wavele | vedclass

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

Question

(A) Total energy provided by the source per second $= \frac{1000 \, J}{10 \, s} = 100 \, J/s$.Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.6

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) Total energy provided by the source per second $= \frac{1000 \, J}{10 \, s} = 100 \, J/s$.Energy of one photon $E = \frac{hc}{\lambda} = \frac{6.626 	imes 10^{-34} \, Js 	imes 3 	imes 10^8 \, m/s}{400 	imes 10^{-9} \, m} = 4.9695 	imes 10^{-19} \, J$.Number of photons incident per second $= \frac{	ext{Total energy per second}}{	ext{Energy of one photon}} = \frac{100}{4.9695 	imes 10^{-19}} \approx 2.012 	imes 10^{20}$.Since one photon ejects one electron,the number of electrons ejected per second is $2.012 	imes 10^{20}$.Comparing this with $x 	imes 10^{20}$,we get $x \approx 2$.

List-$I$	List-$II$
$(1)$ Exclusion Principle	$(A)$ Hund
$(2)$ Multiplicity Rule	$(B)$ Heisenberg
$(3)$ Uncertainty Principle	$(C)$ Einstein
$(4)$ Quantum Theory	$(D)$ Planck
	$(E)$ Pauli

Similar Questions

Write down the postulates of Bohr's model for the hydrogen atom.

What is the difference between a photon and a quantum?

The radius of a hydrogen atom in its ground state is $0.53 \ \mathring{A}$. The radius of the $Li^{2+}$ ion in the same state (atomic number = $3$) is ............. $\mathring{A}$.

Match the principles given in List-$I$ with their discoverers given in List-$II$.

List-$I$ List-$II$

$(1)$ Exclusion Principle $(A)$ Hund

$(2)$ Multiplicity Rule $(B)$ Heisenberg

$(3)$ Uncertainty Principle $(C)$ Einstein

$(4)$ Quantum Theory $(D)$ Planck

$(E)$ Pauli

The binding energy of an $e^-$ in the ground state of a hydrogen atom is $13.6 \ eV$. The energy required to remove an electron from the ground state of a $He^+$ ion is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module