Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths `lambda_(N) ,lambda_(A)` respectively. The ratio `(lambda_(N))/(lambda_(A))` is closest to :
An alpha particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are lambda_(alpha) and lambda_(p) respectively. The ratio (lambda_(p))/(lambda_(alpha)) is :
When radiations of wavelength lambda_(1), lambda_(2) (= 0.6 lambda_(1)), lambda_(3) (= 4lambda_(2)) and lambda_(4) are incident on a metal plate, photoelectrons with maximum kinetic energy of – 5.2 eV are emitted for lambda_(1), 12 eV are emitted for lambda_(2), 0.95 eV are emitted for lambda_(4) and no electron is emitted for l3. It is known that a hydrogen like atom (atomic number Z) in a higher excited state of quantum number n can make a transition to first excited state by successively emitting two photons of wavelength lambda1 and lambda_(2) respectively. Alternatively, the atom from same excited state can make a transition to the second excited state by successively emitting two photons of wavelength lambda_(3) and lambda_(4) respectively. (a) Find the work function of the metal. (b) Find the values of n and Z for the atom.
A dye absorbs a photon of wavelength lambda and re - emits the same energy into two phorons of wavelengths lambda_(1) and lambda_(2) respectively. The wavelength lambda is related with lambda_(1) and lambda_(2) as :
An a- particle and a proton are accelerated from rest by a potential difference of 100 V After this their de Broglie wavelength are lambda_(a) and lambda_(p) respectively The ratio (lambda_(p))/ lambda_(p)) , to the rearest integer is
An a- particle and a proton are accelerated from rest by a potential difference of 100V. After this, their de-Broglie wavelengths are lambda_a and lambda_p respectively. The ratio (lambda_p)/(lambda_a) , to the nearest integer, is.
Two sound waves of wavelengths lambda_(1) and lambda_(2) (lambda _(2) gt lambda_(1)) produces n beats//s , the speed of sound is
Electron in hydrogen atom first jumps from third excited state to second excited state and then form second excited state to first excited state. The ratio of wavelength lambda_(1): lambda_(2) emitted in two cases is
An electron approaches a fixed proton from a large distance with a kinetic energy of 12.2 eV. The electron gets captured by the proton to form a hydrogen atom in an excited state and a photon of wavelength lambda_(1) = 796 Å is emitted. Later, the hydrogen atom de-excites by emitting a single photon of wavelength lambda_(2) . Find lambda_(2) . [Take h = 6.62 xx 10^(– 34) Js, c = 3 xx 108 m//s ]
