Atomic StructureHard
Question
Photon having energy equivalent to the binding energy of 4th state of He+ atom is used to eject an electron from the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of 4V then the minimum value of De-broglie wavelength associated with the electron is :
Options
A.1.1 (Ao)
B.5 (Ao)
C.9.15 (Ao)
D.11 (Ao)
Solution
Total energy = 
= 3.4 eV
Now K.E. = 3.4 - 1.4 = 2 eV
Now, Total energy = 2 + 4 = 6 eV i.e. potential = 6 V
For electron λ =
so λ = 5 (Ao).
= 3.4 eVNow K.E. = 3.4 - 1.4 = 2 eV
Now, Total energy = 2 + 4 = 6 eV i.e. potential = 6 V
For electron λ =
so λ = 5 (Ao).Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
The K and L shell of an element are completely filled and there are 16 electrons in M-shell and 2 electrons in N-shell. ...The correct order for the electronegetivity of N, O, F and P is :-...The angular momentum of electron in Bohr′s orbit is J. What will be the potential energy of that Bohr′s orbi...How many nodal spheres are present in 4s-orbital....The I.P. of an atom is 16 eV. What is the value of first excitation potential :-...