Question 251:
easyThe K.E. of electron and photon is same then relation between their De-Broglie wavelength:
(1999)
Question 251:
easyThe K.E. of electron and photon is same then relation between their De-Broglie wavelength:
(1999)
Question 252:
easyAn electron beam has a kinetic energy equal to 100 eV. Find its wavelength associated with a beam, if mass of electron = $9.1 \times 10^{-31} kg$ and $1 eV = 1.6 \times 10^{-19} J/eV$. (Planck’s constant = $6.6 \times 10^{-34} Js$).
(1996)
Question 253:
easyAn electron of mass m, when accelerated through a potential difference V, has de-Broglie wavelength $\lambda$. The de-Broglie wavelength associated with a proton of mass M accelerated through the same potential difference, will be
(1995)
Question 254:
easyThe de Broglie wavelength of an electron moving with kinetic energy of 144 eV is nearly,
(2020-Covid)
Question 255:
easyAn electron is accelerated through a potential difference of 10,000 V. Its de Broglie wavelength is, (nearly) : ($m_e = 9 \times 10^{-31} kg$)
(2019)
Question 256:
easyAn electron of mass m with an initial velocity $\vec{v} = v_0\hat{i} (v_0 > 0)$ enters an electric field $\vec{E} = -E_0\hat{i} (E_0 = constant > 0)$ at t = 0. If $\lambda_0$ is its de-Broglie wavelength initially, then its de-Broglie wavelength at time t is
(2018)
Question 257:
easyThe de-Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature T (Kelvin) and mass m, is:
(2017-Delhi)
Question 258:
easyAn electron of mass m and a photon have same energy E. The ratio of de-Broglie wavelengths associated with them is (c being velocity of light)
(2016-I)
Question 259:
easyElectrons of mass m with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength ($\lambda_0$) of the emitted X-ray is:
(2016-II)
Question 260:
easyIf a photon has velocity c and frequency $\nu$, then which of the following represents its wavelength?
(1996)
The energy of a photon is $E = \frac{hc}{\lambda}$. Rearranging for wavelength gives $\lambda = \frac{hc}{E}$.